Class 11

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 13 Limits and Derivatives Ex 13.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 13 Limits and Derivatives Ex 13.1

Question 1.
Evaluate the given limit: \(\lim _{x \rightarrow 3}\) x + 3.
Answer.
\(\lim _{x \rightarrow 3}\) x + 3 = 3 + 3 = 6.

Question 2.
Evaluate the given limit: \(\lim _{x \rightarrow \pi}\) (x – \(\frac{22}{7}\)).
Answer.
\(\lim _{x \rightarrow \pi}\) (x – \(\frac{22}{7}\)) = (π – \(\frac{22}{7}\)).

Question 3.
Evaluate the given limit : \(\lim _{x \rightarrow 1}\) πr².
Answer.
\(\lim _{x \rightarrow 1}\) πr² = π (1)² = π.

Question 4.
Evaluate the given limit : \(\lim _{x \rightarrow 4} \frac{4 x+3}{x-2}\).
Answer.
\(\lim _{x \rightarrow 4} \frac{4 x+3}{x-2}=\frac{4(4)+3}{4-2}=\frac{16+3}{2}=\frac{19}{2}\)

Question 5.
Evaluate the given limit: \(\lim _{x \rightarrow-1} \frac{x^{10}+x^{5}+1}{x-1}\)
Answer.
\(\lim _{x \rightarrow-1} \frac{x^{10}+x^{5}+1}{x-1}=\frac{(-1)^{10}+(-1)^{5}+1}{-1-1}\)

= \(\frac{1-1+1}{-2}=-\frac{1}{2}\)

Question 6.
Evaluate the given limit : \(\lim _{x \rightarrow 0} \frac{(x+1)^{5}-1}{x}\).
Answer.

Question 7.
Evaluate the given limit : \(\lim _{x \rightarrow 2} \frac{3 x^{2}-x-10}{x^{2}-4}\).
Answer.
At x = 2, the value of the given rational function takes the form \(\frac{0}{0}\).
∴ \(\lim _{x \rightarrow 2} \frac{3 x^{2}-x-10}{x^{2}-4}=\lim _{x \rightarrow 2} \frac{(x-2)(3 x+5)}{(x-2)(x+2)}\)

\(\lim _{x \rightarrow 2} \frac{3 x+5}{x+2}=\frac{3(2)+5}{2+2}=\frac{11}{4}\)

Question 8.
Evaluate the given limit: \(\lim _{x \rightarrow 3} \frac{x^{4}-81}{2 x^{2}-5 x-3}\).
Answer.
At x = 3, the value of the given rational function takes the form \(\frac{0}{0}\).
∴ \(\lim _{x \rightarrow 3} \frac{x^{4}-81}{2 x^{2}-5 x-3}\) = \(\lim _{x \rightarrow 3} \frac{(x-3)(x+3)\left(x^{2}+9\right)}{(x-3)(2 x+1)}\)

\(\lim _{x \rightarrow 3} \frac{(x+3)\left(x^{2}+9\right)}{2 x+1}\) = \(\frac{(3+3)\left(3^{2}+9\right)}{2(3)+1}\)

= \(\frac{6 \times 18}{7}=\frac{108}{7}\)

Question 9.
Evaluate the given limit; \(\lim _{x \rightarrow 0} \frac{a x+b}{c x+1}\).
Answer.
\(\lim _{x \rightarrow 0} \frac{a x+b}{c x+1}=\frac{a(0)+b}{c(0)+1}\) = .

Question 10.
Evaluate the given limit: \(\lim _{z \rightarrow 1} \frac{z^{\frac{1}{3}}-1}{z^{\frac{1}{6}}-1}\).
Answer.

Question 11.
Evaluate the given limit: \(\lim _{x \rightarrow 1} \frac{a x^{2}+b x+c}{c x^{2}+b x+a}\), a + b + c ≠ 0.
Answer.
\(\lim _{x \rightarrow 1} \frac{a x^{2}+b x+c}{c x^{2}+b x+a}=\frac{a(1)^{2}+b(1)+c}{c(1)^{2}+b(1)+a}\)

= \(\frac{a+b+c}{a+b+c}\) = 1. [a + b + c ≠ 0]

Question 12.
Evaluate the given limit: \(\lim _{x \rightarrow-2} \frac{\frac{1}{x}+\frac{1}{2}}{x+2}\)
Answer.

Question 13.
Evaluate the given limit: \(\lim _{x \rightarrow 0} \frac{\sin a x}{b x}\).
Answer.

Question 14.
Evaluate the given limit: \(\lim _{x \rightarrow 0} \frac{\sin a x}{\sin b x}\), a, b ≠ 0.
Answer.

Question 15.
Evaluate the given limit: \(\lim _{x \rightarrow \pi} \frac{\sin (\pi-x)}{\pi(\pi-x)}\)
Answer.
\(\lim _{x \rightarrow \pi} \frac{\sin (\pi-x)}{\pi(\pi-x)}\)
put π – x = θ, As x → π, θ → 0 (zero)
\(\lim _{\theta \rightarrow 0} \frac{\sin \theta}{\pi \theta}=\lim _{\theta \rightarrow 0} \frac{1}{\pi} \frac{(\sin \theta)}{\theta}=\frac{1}{\pi}\)

Question 16.
Evaluate the given limit: \(\lim _{x \rightarrow 0} \frac{\cos x}{\pi-x}\).
Answer.
\(\lim _{x \rightarrow 0} \frac{\cos x}{\pi-x}=\frac{\cos 0}{\pi-0}=\frac{1}{\pi}\)

Question 17.
Evaluate the given limit: \(\lim _{x \rightarrow 0} \frac{\cos 2 x-1}{\cos x-1}\).
Answer.

Question 18.
Evaluate the given limit: \(\lim _{x \rightarrow 0} \frac{a x+x \cos x}{b \sin x}\)
Answer.

Question 19.
Evaluate the given limit: \(\lim _{x \rightarrow 0}\) x sec x.
Answer.
\(\lim _{x \rightarrow 0}\) x sec x = \(\lim _{x \rightarrow 0} \frac{x}{\cos x}=\frac{0}{\cos 0}=\frac{0}{1}\) = 0.

Question 20.
Evaluate the given limit: \(\) a, b, a + b ≠ 0.
Answer.

Question 21.
Evaluate the given limit: \(\lim _{x \rightarrow 0}\) (cosec x – cot x).
Answer.
\(\lim _{x \rightarrow 0}\) (cosec x – cot x)
At x = 0, the value of the given function takes the form ∞ – ∞.

Question 22.
Evaluate the given limit \(\lim _{x \rightarrow \frac{\pi}{2}} \frac{\tan 2 x}{x-\frac{\pi}{2}}\).
Answer.
\(\lim _{x \rightarrow \frac{\pi}{2}} \frac{\tan 2 x}{x-\frac{\pi}{2}}\)

At x = \(\frac{\pi}{2}\), the value of the given function takes the form \(\frac{0}{0}\).

Question 23.
Find \(\lim _{x \rightarrow 0}\) f(x) and \(\lim _{x \rightarrow 1}\) f(x), where f(x) = .
Answer.
Given, f(x) =
(i) Now, LHL = \(\lim _{x \rightarrow 0^{-}}\) f(x) = \(\lim _{x \rightarrow 0^{-}}\) (2x + 3)
= \(\lim _{h \rightarrow 0}\) [2 (0 – h) + 3] = 3
[putting x = 0 – h as x → 0, then h → 0]
RHL = \(\lim _{x \rightarrow 0^{+}}\) f(x) = \(\lim _{x \rightarrow 0^{+}}\) 3(x + 1)
= \(\lim _{h \rightarrow 0}\) [3(0 + h) + 1] = 3
[putting x = 0 + h as x → 0,then h → 0]J
Here, LHL = RHL
∴ \(\lim _{x \rightarrow 0}\) f(x) = 3

(ii) We have to find \(\lim _{x \rightarrow 1}\) f(x)
\(\lim _{x \rightarrow 1}\) f(x) = \(\lim _{x \rightarrow 1}\) 3 (x + 1)
= 3 (1 + 1) = 6

Question 24.
Find \(\lim _{x \rightarrow 1}\) f(x), where f(x) =
Answer.
Given, f(x) =
At x = 1,
RHL = \(\lim _{x \rightarrow 1^{+}}\) f(x) = \(\lim _{h \rightarrow 0}\) f(1 + h)
= \(\) (1 + h)² – 1 [put x = 1 + h]
= – (1 + 0)² – 1
= – 1 – 1 = – 2
LHL = \(\lim _{x \rightarrow 1^{-}}\) f(x) = \(\lim _{h \rightarrow 0}\) f(1 – h)
= \(\lim _{h \rightarrow 0}\) (1 – h)² – 1 [put x = 1 – h]
= (1 – 0)² – 1 = 1 – 1 = 0
RHL ≠ LHL
Hence, at x = 1 , limit does not exist.

Question 25.
Evaluate \(\lim _{x \rightarrow 0}\) f(x), where f(x) = .
Answer.

Question 26.
Find \(\lim _{x \rightarrow 0}\) f(x), where f(x) = .
Answer.

Question 27.
Find \(\lim _{h \rightarrow 5}\) f(x), where f(x) = |x| – 5.
Answer.
The given function is f(x) = |x| – 5
when x > 5, put x = 5 + h, where h is small
|x| = |5 + h| = 5 + h
∴ \(\lim _{x \rightarrow 5^{+}}\) f(x) = \(\lim _{h \rightarrow 0}\) [(5 + h) – 5] = \(\lim _{h \rightarrow 0}\) h = 0
when x < 5, put x = 5 – h, where h is small
∴ |5 – h| = 5 – h
∴ \(\lim _{x \rightarrow 5^{-}}\) f(x) = \(\lim _{h \rightarrow 0}\) (5 – h – 5) = \(\lim _{h \rightarrow 0}\) (- h) = 0
∴ \(\lim _{x \rightarrow 5^{-}}\) f(x) = \(\lim _{x \rightarrow 5^{+}}\) f(x) = 0
∴ \(\lim _{h \rightarrow 5}\) f(x) = 0

Question 28.
Suppose f(x) = and if \(\lim _{h \rightarrow 1}\) f(x) = f(1) what are possible values of a and b ?
Answer.
The given function is f(x) =
\(\lim _{x \rightarrow 1^{-}}\) f(x) = \(\lim _{x \rightarrow 1}\) (a + bx) = a + b
\(\lim _{x \rightarrow 1^{-}}\) f(x) = \(\lim _{x \rightarrow 1}\) (b – ax) = b – a
f(1) = 4
It is given that \(\lim _{x \rightarrow 1}\) f(x) = f(1).
∴ \(\lim _{x \rightarrow 1^{-}}\) f(x) = \(\lim _{x \rightarrow 1^{+}}\) f(x)
= \(\lim _{x \rightarrow 1}\) f(x) = f(1)
a + b = 4 and b – a = 4.
On solving these two equations, we obtain a = 0 and b = 4.
Thus, the respective possible values of a and b are 0 and 4.

Question 29.
Let a₁, a₂, ………. a, be fixed real numbers and define a function f(x) = (x – a₁) (x – a₂) ………….. (x – a_n). What is \(\lim _{x \rightarrow a_{1}}\) f(x)? For some a ≠ a₁, a₂, ………………. a_n compute \(\lim _{x \rightarrow a}\) f(x).
Answer.

Question 30.
If f(x) = For what value(s) of a does f(x) exists?
Answer.

Question 31.
If the function f(x) satisfies \(\lim _{x \rightarrow 1} \frac{f(x)-2}{x^{2}-1}\) = π, evaluate \(\lim _{x \rightarrow 1}\) f(x).
Answer.

Question 32.
If f(x) = For what integers m and n does \(\lim _{x \rightarrow 0}\) f(x) and \(\lim _{x \rightarrow 1}\) f(x) exist?
Answer.

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 13 Limits and Derivatives Ex 13.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 13 Limits and Derivatives Ex 13.2

Question 1.
Find the derivative of x² – 2 at x = 10.
Answer.
We have, f(x) = x² – 2
By using first principle of derivative,
f'(10) = \(\lim _{h \rightarrow 0} \frac{f(10+h)-f(10)}{h}\)

= \(\lim _{h \rightarrow 0} \frac{\left[(10+h)^{2}-2\right]-\left[(10)^{2}-2\right]}{h}\)

= \(\lim _{h \rightarrow 0} \frac{100+20 h+h^{2}-2-100+2}{h}\)

= \(\lim _{h \rightarrow 0} \frac{h^{2}+20 h}{h}\)

= \(\lim _{h \rightarrow 0}\) (h + 20) = 20.

Question 2.
Find the derivative of 99x at x = 100.
Answer.
Let f(x) = 99x. Accordingly,
f'(100) = \(\lim _{h \rightarrow 0} \frac{f(100+h)-f(100)}{h}\)

= \(\lim _{h \rightarrow 0} \frac{99(100+h)-99(100)}{h}\)

= \(\lim _{h \rightarrow 0} \frac{99 \times 100+99 h-99 \times 100}{h}\)

= \(\lim _{h \rightarrow 0} \frac{99 h}{h}\)

= \(\lim _{h \rightarrow 0}\) (99) = 99
Thus, the derivative of 99x at x = 100 is 99.

Question 3.
Find the derivative of x at x = 1.
Answer.
Let f(x) = x. Accordingly,
f(1) = \(\lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{h}\)

= \(\lim _{h \rightarrow 0} \frac{(1+h)-1}{h}=\lim _{h \rightarrow 0} \frac{h}{h}=\lim _{h \rightarrow 0}(1)\) = 1.
Thus, the derivative of x at x = 1 is 1.

Question 4.
Find the derivative of the following functions from first principle.
(i) x³ – 27
(ii) (x – 1) (x – 2)
(iii) \(\frac{1}{x^{2}}\)
(iv) \(\frac{x+1}{x-1}\)
Answer.
(i) We have, f(x) = x³ – 27
By using first principle of derivative,

(ii) We have, f(x) = (x – 1) (x – 2) = x² – 3x + 2
By first principle of derivative, we have

(iii) Let f(x) = \(\frac{1}{x^{2}}\).
Accordingly, from the first, principle,

(iv) Let f(x) = \(\frac{x+1}{x-1}\)
Accordingly, from the first principle,

Question 5.
For the function f(x) = \(\). Prove that f'(1) = 100 f'(0).
Answer.

At x = 0, f'(0) = 1
At x = 1, f'(1) = 1⁹⁹ + 1⁹⁸ + ………….. + 1 + 1
= [1 + 1 + 1 + …………… + 1 + 1]_{100 terms}
= 1 × 100 = 100.

Question 6.
Find the derivative of xⁿ + ax^{n – 1} + a² x^{n – 2} + ………….. + a^{n – 1} x + aⁿ for some fixed real number a.
Answer.
Let f(x) = xⁿ + ax^{n – 1} + a² x^{n – 2} + ………….. + a^{n – 1} x + aⁿ
On differentiating both sides, we get
f'(x) = nx^{n – 1} + a(n – 1)x^{n – 2} + a² (n – 2)x^{n – 3} + …………. + a^{n – 1} . 1 + 0
On putting x = a both sides, we get
f'(a) = na^{n – 1} + a (n – 1) x^{n – 2} + a² (n – 2) a^{n – 3} + ……………. + a^{n – 1}
= n a^{n – 1} + (n – 1) a^{n – 1} + (n – 2) a^{n – 1} + ……………. + a^{n – 1}
= a^{n – 1} [n + (n – 1) + (n – 2) + ………….. + 1]
[∵ sum of n natural numbers = \(\frac{n(n+1)}{2}\)]
= \(a^{n-1} \frac{n(n+1)}{2}\).

Question 7.
For some constants a and b, find the derivative of
(i) (x – a) (x – b)
(ii)(ax +b)
(iii) \(\frac{x-a}{x-b}\)
Ans.
(i) Let f(x) = (x – a) (x – b)
f(x)= x² – (a + b) x + ab
∴ f'(x) = \(\frac{d}{d x}\) [x² – (a + b) x + ab]
= \(\frac{d}{d x}\) (x²) – (a + b) \(\frac{d}{d x}\)(x) + \(\frac{d}{d x}\) (ab)
On using theorem \(\frac{d}{d x}\) (xⁿ) = nx^{n – 1}, we obtain
f'(x) = 2x – (a + b) + 0 = 2x – a – b.

(ii) Let f(x) = (ax² + b)²
f(x) = a²x⁴ + 2abx² + b²
∴ f'(x) = \(\frac{d}{d x}\) (a²x⁴ + 2abx² + b²)
= a² \(\frac{d}{d x}\) (x⁴) + 2ab \(\frac{d}{d x}\) (x²) + \(\frac{d}{d x}\) (b²)
On using theorem \(\frac{d}{d x}\) (xⁿ) = nx^{n – 1}, we obtain
f’(x) = a² (4x³) + 2ab (2x) + b2(O) = 4a2x3 + 4abx= 4ax (ax2 + b).

(iii) Let f(x) = \(\frac{x-a}{x-b}\)

Question 8.
Find the derivative of \(\frac{x^{n}-a^{n}}{x-a}\) for some constant a.
Answer.

Question 9.
Find the derivative of
(i) 2x – \(\frac{3}{4}\)
(ii) (5x³ + 3x – 1) (x – 1)
(iii) x^{– 3} (5 + 3x)
(iv) x⁵ (3 – 6x^{– 9})
(v) x^-4 (3 – 4x^{– 5})
(vi) \(\frac{2}{x+1}-\frac{x^{2}}{3 x-1}\)
Answer.
(i) Let f(x) = 2x – \(\frac{3}{4}\)
f'(x) = \(\frac{d}{d x}\left(2 x-\frac{3}{4}\right)\)

= \(2 \frac{d}{d x}(x)-\frac{d}{d x}\left(\frac{3}{4}\right)\)

= 2 – 0 = 2

(ii) Let f(x) = (5x³ + 3x – 1) (x – 1)
By Leibnitz product rule,
f(x) = (5x³ + 3x – 1) \(\frac{d}{d x}\) (x -1) + (x – 1) \(\frac{d}{d x}\) (5x³ + 3x – 1)
= (5x³ + 3x – 1) (1) + (x – 1) (53x² + 3 – 0)
= (5x³ +3x – 1) +(x – 1) (15x² + 3)
= 5x³ + 3x – 1 + 15x³3 + 3x – 15x² – 3
= 20x³ – 15x² + 6x – 4.

(iii) Let f(x) = x^{– 3} (5 + 3x) .
By Leibnitz product rule,
f(x) = x^{– 3} \(\frac{d}{d x}\) (5 + 3x) + (5 + 3x) \(\frac{d}{d x}\) (x^{– 3})
= x^{– 3} (0 + 3) + (5 + 3x) (- 3x^{– 3 – 1})
= x^{– 3} (3) + (5 + 3x) (- 3x^{– 4})
= – 3x^{– 3} (2 + \(\frac{5}{x}\))
= \(\frac{-3 x^{-3}}{x}\) (2x + 5)
= \(=\frac{-3}{x^{4}}\) (5 + 2x)

(iv) Let f(x) = x⁵ (3 – 6x^-9)
By Leibnitz product rule,
f'(x) = \(x^{5} \frac{d}{d x}\left(3-6 x^{-9}\right)+\left(3-6 x^{-9}\right) \frac{d}{d x}\left(x^{5}\right)\)
= x⁵ {0 – 6 (- 9) x^{– 9 – 1}} + (3 – 6x^{– 9}) (5x⁴)
= x⁵ (54x^{– 10}) + 15x⁴ – 30x^{– 5}
= 54x^{– 5} + 15x⁴ – 30x^{– 5}
= 24x^{– 5} + 15x⁴
= 15x⁴ + \(\frac{24}{x^{5}}\)

(v) Let f(x) = x^{– 4} (3 – 4x^{– 5})
By Leibnitz product rule,
f’(x) = x^{– 4} \(\frac{d}{d x}\) (3 – 4x^{– 5}) + (3 – 4x^{– 5}) \(\frac{d}{d x}\) (x^{– 4})
= x^{– 4} {0 – 4 (- 5) x^{– 5 – 1}} + (3 – 4x^{– 5}) (- 4)x^{– 4 – 1}
= x^{– 4} (20 x^{– 6}) + (3 – 4x^{– 5}) (- 4 x^{– 5})
= 20x^{– 10} – 12x^{– 5} + 16x^{– 10}
= 36x^{– 10} – 12x^{– 5}
= \(-\frac{12}{x^{5}}+\frac{36}{x^{10}}\).

(vi)

Question 10.
Find the derivative of cos x from first principle.
Answer.
Let f(x) = cos x.
Accordingly, from the first principle, /'(*) = Um /(* +10-/00 = ^ COSUM 111 <mx

= – cos x (0) – sin x(1)
[∵ \(\lim _{h \rightarrow 0} \frac{1-\cos h}{h}\) = 0 and \(\lim _{h \rightarrow 0} \frac{\sin h}{h}\) = 1]
= – sin x
∴ f'(x) = – sin x

Question 11.
Find the derivative of the following functions:
(i) sin x cos x
(ii) sec x
(iii) 5 sec x + 4 cos x
(iv) cosec x
(v) 3 cot x + 5 cosec x
(vi) 5 sin x – 6 cos x+7
(vii) 2 tan x – 7 sec x
Answer.
(i) Let f(x) = sin x cos x.
Accordingly, from the first principle,

(ii) Let (x) = sec x.
Accordingly, from the first principle.

= \(\frac{1}{\cos x} \lim _{\frac{h}{2} \rightarrow 0} \frac{\sin \left(\frac{h}{2}\right)}{\left(\frac{h}{2}\right)} \lim _{h \rightarrow 0} \frac{\sin \left(\frac{2 x+h}{2}\right)}{\cos (x+h)}\)

= \(\frac{1}{\cos x} \cdot 1 \cdot \frac{\sin x}{\cos x}\)

= sec x tan x.

(iii) Let f(x) = 5 sec x + 4 cos x.
Accordingly, from the first principle,

= \(\frac{5}{\cos x}\left[\lim _{h \rightarrow 0} \frac{\sin \left(\frac{2 x+h}{2}\right)}{\cos (x+h)} \lim _{\frac{h}{2} \rightarrow 0} \frac{\sin \left(\frac{h}{2}\right)}{\frac{h}{2}}\right]\) -4 sin x

= \(\frac{5}{\cos x} \cdot \frac{\sin x}{\cos x} \cdot 1\) – 4 sin x

= 5 sec x tan x – 4 sin x

(iv) Let f(x) = cosec x
Accordingly, from the first principle,

(v) Let f(x) = 3cot x – 5 cosec x
Accordingly, from the first principle,

(vi) Let f(x) = 5sin x – 6cos x + 7.
Accordingly, from the first principle,
f’(x) = \(\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\)
= \(\lim _{h \rightarrow 0} \frac{1}{h}\) [5 sin (x + h) – 6 cos (x + h) + 7 – 5 sin x + 6 cos x – 7]

= \(\lim _{h \rightarrow 0} \frac{1}{h}\) [5 {sin (x + h) – sin x) – 6 {cos (x + h) – cos x}]

= 5 \(\lim _{h \rightarrow 0} \frac{1}{h}\) [sin(x + h) – sin x] – 6 \(\lim _{h \rightarrow 0} \frac{1}{h}\) [cos(x + h) – cos x]

= \(5 \lim _{h \rightarrow 0} \frac{1}{h}\left[2 \cos \left(\frac{x+h+x}{2}\right) \sin \left(\frac{x+h-x}{2}\right)\right]\) – 6 \(\lim _{h \rightarrow 0} \frac{\cos x \cos h-\sin x \sin h-\cos x}{h}\)

= 5 \(\lim _{h \rightarrow 0} \frac{1}{h}\left[2 \cos \left(\frac{(2 x+h)}{2}\right) \sin \frac{h}{2}\right]\) – 6 \(\lim _{h \rightarrow 0} \frac{1}{h}\left[\frac{-\cos x(1-\cos h)-\sin x \sin h}{h}\right]\)

= 5 cos x . 1 – 6 [(- cos x) . (0) – sin x . 1]
= 5 cos x + 6 sin x

(vi) Let f(x) = 2 tan x – 7 sec x
Accordingly, from the first principle,

= \(\) \(\)

= \(\)

= 2 sec² x – 7 sec x tan x.

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 12 Introduction to Three Dimensional Geometry Miscellaneous Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry Miscellaneous Exercise

Question 1.
Three vertices of a parallelogram ABCD are A (3,- 1, 2), B(1, 2, – 4) and C (- 1, 1, 2). Find the coordinates of the fourth vertex.
Answer.
The three vertices of a parallelogramABCD are given as A (3,- 1, 2), B (1, 2, – 4) and C (- 1, 1, 2).
Let the coordinates of the fourth vertex be D(x, y, z).

We know that the diagonals of a parallelogram bisect each other.
Therefore, in parallelogram ABCD, AC and BD bisect each other.
∴ Mid-point of AC = Mid-point of BD
\(\left(\frac{3-1}{2}, \frac{-1+1}{2}, \frac{2+2}{2}\right)=\left(\frac{x+1}{2}, \frac{y+2}{2}, \frac{z-4}{2}\right)\)
(1, 0, 2) = \(\left(\frac{x+1}{2}, \frac{y+2}{2}, \frac{z-4}{2}\right)\)
\(\frac{x+1}{2}\) = 1, \(\frac{y+2}{2}\) = 0, and \(\frac{z-4}{2}\) = 2
x = 1 ,y = – 2, and z = 8
Thus, the coordinates of the fourth vertex are (1, – 2, 8).

Question 2.
Find the lengths of the medians of the trisìngle with vertices A (0, 0, 6), B (0, 4, 0) and C (6, 0, 0).
Answer.
ABC is a thangle with vertices A (0, 0, 6), B (0, 4, 0) and C (6, 0, 0).

[∵ Coordinates of mid-points = \(\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}, \frac{z_{1}+z_{2}}{2}\right)\)]

Let points D, E and F are the mid-points of BC, AC and AB, respectively.
So, AD, BE and CF will be the medians of the triangle.
Coordinates of point D = \(\left(\frac{0+6}{2}, \frac{4+0}{2}, \frac{0+0}{2}\right)\) = (3, 2, 0)

Coordinates of point E = \(\left(\frac{0+6}{2}, \frac{0+0}{2}, \frac{6+0}{2}\right)\) = (3, 0, 3)

Coordinates of point F = \(\left(\frac{0+0}{2}, \frac{0+4}{2}, \frac{6+0}{2}\right)\) = (0, 2, 3)

Thus, the coordinates of the fourth vertex are (1, – 2, 8).

Now, length of median
AD = Distance between points A and D
AD = \(\sqrt{(0-3)^{2}+(0-2)^{2}+(6-0)^{2}}\)
[∵ distance = \(\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}+\left(z_{1}-z_{2}\right)^{2}}\)]
= \(\sqrt{9+4+36}\)
= √49 = 7

similarly, BE = \(\sqrt{(0-3)^{2}+(4-0)^{2}+(0+3)^{2}}\)
= \(\sqrt{9+16+9}\) = √34

and CF = \(\sqrt{(6-0)^{2}+(0-2)^{2}+(0-3)^{2}}\)
= \(\sqrt{36+4+9}\) = √49 = 7
Hence, length of the median are 7, √34 and 7.

Question 3.
If the origin is the centroid of the triangle PQR with vertices P (2a, 2, 6), Q (- 4, 3b, -10) and R (8, 14, 2c), then find the values of a, b and c.
Answer.

It is known that the coordinates of the centroid of the triangle, whose vertices are (x₁, y₁, z₁), (x₂, y₂, z₂) and (x₃, y₃, z₃),are \(\left(\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}, \frac{z_{1}+z_{2}+z_{3}}{3}\right)\)

Therefore, coordinates of the centroid of ∆PQR = \(\left(\frac{2 a-4+8}{3}, \frac{2+3 b+14}{3}, \frac{6-10+2 c}{3}\right)\)

= \(\left(\frac{2 a+4}{3}, \frac{3 b+16}{3}, \frac{2 c-4}{3}\right)\)

It is given that the origin is the centroid of ∆PQR.
∴ (0, 0, 0) = \(\left(\frac{2 a+4}{3}, \frac{3 b+16}{3}, \frac{2 c-4}{3}\right)\)

\(\frac{2 a+4}{3}\) = 0, \(\frac{3 b+16}{3}\) = 0 and \(\frac{2 c-4}{3}\) = 0

a = – 2, b = – \(\frac{16}{3}\) and c = 2.

Thus, the respective values of a, b and c are – 2, – \(\frac{16}{3}\) and 2.

Question 4.
Find the coorlinsites of a point on y-axis which are at a distance of 5√2 from point P(3, – 2, 5).
Answer.
If a point is on the y-axis, then x-coordinate and the z-coordinate of the point are zero.
Let A (0, b, 0) be the point on the y-axis at a distance of 5√2 from point P (3, – 2, 5).
Accordingly, AP = 5√2
∴ AP² = 50
(3 – 0)² + (- 2 – b)² + (5 – 0)² = 50
⇒ 9 + 4 + b² + 4b + 25 = 50
⇒ b² + 4b – 12 = 0
⇒ b² + 6b – 2b – 12 = 0
⇒ (b + 6) (b – 2) = 0
⇒ b = – 6 o r2.
Thus, the coordinates of the required points are (0, 2, 0) and (0, – 6, 0).

Question 5.
A point R with x-coordinnte 4 lies on the line segment joining the points P (2, – 3, 4) and Q(8, 0, 10). Find the coordinates of the point R.
[Hint : Suppose R divides PQ in the ratio k : 1. The co-ordinates of the point R are given by, \(\left(\frac{8 k+2}{k+1}, \frac{-3}{k+1}, \frac{10 k+4}{k+1}\right)\)]
Answer.
Let the coordinates of R be (4, y, z) and R divides PQ in ratio k : 1
∴ Coordinate of R is \(\left(\frac{8 k+2}{k+1}, \frac{-3}{k+1}, \frac{10 k+4}{k+1}\right)\).
[∵ using internal ratio formula]

But x – coordinate of R is 4.
So, \(\frac{8 k+2}{k+1}\) = 4
⇒ 8k + 2 = 4k + 4
⇒ 4k = 2
⇒ k = \(\frac{1}{2}\)

img 5

Hence, coordinates of R are (4, – 2, 6).

Question 6.
If A and B be the points (3, 4, 5) and (- 1, 3, – 7), respectively, find the equation of the set of points P such that PA² + PB² = k², where k is a constant.
Answer.
The coordinates of points A and B are given as (3, 4, 5) and Q (- 1, 3, – 7) respectively.
Let the coordinates of point P be (x, y, z).
On using distance formula, we obtain .
PA² = (x – 3)² + (y – 4)² + (z – 5)²
= x² + 9 – 6x + y² +16 – 8y + z² + 25 – 10z
= x² – 6x + y² – 8y + z² – 10z + 50
PB² = (x + 1)² + (y – 3)² + (z + 7)²
= x² + 2x +1 + y² – 6y + 9 + z² + 14z + 49
= x² + 2x + y² – 6y + z² + 14z + 59
Now, if PA² + PB² = k², then
(x² – 6x + y² – 8y + z² – 10z + 50) + (x² + 2x + y² – 6y + z² + 14z + 59) = k²
⇒ 2x² + 2y² + 2z² – 4x – 14y + 4z +109 = k²
⇒ 2(x² + y² + z² – 2x – 7y + 2z) = k² – 109
⇒ x² + y² + z² – 2x – 7y + 2z = \(\frac{k^{2}-109}{2}\)
Thus, the required equation is x² + y² + z² – 2x + 7y + 2z = \(\frac{k^{2}-109}{2}\).

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 12 Introduction to Three Dimensional Geometry Ex 12.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry Ex 12.3

Question 1.
Find the coordinates of the point which divides the line segment joining the points (- 2, 3, 5) and (1, – 4, 6) in the ratio (i) 2 : 3 internally,
(ii) 2 : 3 externally.
Answer.
Let P(x, y, z) be any point which divides the line segment joining points A(- 2, 3, 5) and B(1, – 4, 6) in the ratio 2 : 3 internally.

Here, the ratio is 2 : 3
m = 2, n = 3
The coordinates of point P = \(\left[\left(\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}}, \frac{m_{1} y_{2}+m_{2} y_{1}}{m_{1}+m_{2}}, \frac{m_{1} z_{2}+m_{2} z_{1}}{m_{1}+m_{2}}\right)\right]\)

= \(\left[\frac{2 \times 1+3 \times(-2)}{2+3}, \frac{2(-4)+3(3)}{2+3}, \frac{2(6)+3(5)}{2+3}\right]\)

= \(\left(\frac{2-6}{5}, \frac{-8+9}{5}, \frac{12+15}{5}\right)=\left(\frac{-4}{5}, \frac{1}{5}, \frac{27}{5}\right)\)

(ii) Let P(x, y, z) be any point which divides the line segment joining points A(- 2, 3, 5) and 8(1, – 4, 6) in the ratio 2 : 3 externally.

Here, the ratio is 2 : 3
∴ m = 2, n = 3
The coordinates of point P = \(\left[\left(\frac{m_{1} x_{2}-m_{2} x_{1}}{m_{1}-m_{2}}, \frac{m_{1} y_{2}-m_{2} y_{1}}{m_{1}-m_{2}}, \frac{m_{1} z_{2}-m_{2} z_{1}}{m_{1}-m_{2}}\right)\right]\)

= \(\left[\frac{2(1)+(-3)(-2)}{2+(-3)}, \frac{2(-4)+(-3) \times 3}{2+(-3)}, \frac{2(6)+(-3)(5)}{2+(-3)}\right]\)

= \(\left(\frac{2+6}{2-3}, \frac{-8-9}{2-3}, \frac{12-15}{2-3}\right)=\left(\frac{8}{-1}, \frac{-17}{-1}, \frac{-3}{-1}\right)\)

= (- 8, 17, 3).

Qiestion 2.
Given that P (3, 2, -4), Q (5, 4, – 6) and R (9, 8, – 10) are coimear. Find the ratio in which Q divides PR.
Answer.
Let point Q(5, 4, -6) divide the line segment joining point P (3, 2, – 4) and R (9, 8, – 10) in the ratio k : 1.
Therefore, by section formula:
(5, 4, – 6) = \(\left(\frac{k(9)+3}{k+1}, \frac{k(8)+2}{k+1}, \frac{k(-10)-4}{k+1}\right)\)

⇒ \(\frac{9 k+3}{k+1}\) = 5
⇒ 9k + 3 = 5k + 5
⇒ 4k = 2
⇒ k = \(\frac{2}{4}=\frac{1}{2}\).
Thus, point Q divides PR in the ratio 1 : 2.

Qiestion 3.
Find the ratio in which the YZ-plane divides the line segment formed by joining the points (- 2, 4, 7) and (3, – 5, 8).
Answer.
Let the YZ plane divide the line segment joining points (- 2, 4, 7) and (3, – 5, 8) in the ratio k: 1.
Hence, by section formula, the coordinates of point of intersection are given by
\(\left(\frac{k(3)-2}{k+1}, \frac{k(-5)+4}{k+1}, \frac{k(8)+7}{k+1}\right)\)

On the YZ plane, the x-coordinate of any point is zero.
⇒ \(\frac{3 k-2}{k+1}\) = 0
⇒ 3k – 2 = 0
⇒ k = \(\frac{2}{3}\)
Thus, the YZ plane divides the line segment formed by joining the given points in the ratio 2 : 3.

Qiestion 4.
Using section formula, show that the points A (2, – 3, 4), B(- 1, 2, 1) and C(0, \(\frac{1}{3}\), 2) are collinear.
Answer.
Let C(0, \(\frac{1}{3}\), 2) divides the join of A(2, – 3, 4) and B(- 1, 2,1) in the ratio k : 1.

Then, coordinates of C are \(\left(\frac{-k+2}{k+1}, \frac{2 k-3}{k+1}, \frac{k+4}{k+1}\right)\) ………………. (i)
[using internal ratio formula]

But coordinates of C are (0, \(\frac{1}{3}\), 2) ………………(ii) [given]

\(\frac{-k+2}{k+1}\) = 0
⇒ – k + 2 = 0
⇒ k = 2

\(\frac{2 k-3}{k+1}=\frac{1}{3}\)
⇒ 6k – 9 = k + 1
⇒ k = 2

\(\frac{k+4}{k+1}\) = 2
⇒ k + 4 = 2k + 2
⇒ k = 2

From each of these equations, we get k = 2
Since, each of these equations give same value of k.
Therefore, the given points are collinear and C divides AB internally in the ratio 2 : 1.

Question 5.
Find the coordinates of the points which trisect the line segment joining the points P(4,2, -6) and Q (10, -16, 6).
Answer.
Let R and S be two points which trisect the joining of P and Q.

Point R divides the join of PQ in the ratio 1 : 2.
∴ Coordinates of R are \(\left[\frac{1(10)+2(4)}{1+2}, \frac{1(-16)+2(2)}{1+2}, \frac{1(6)+2(-6)}{1+2}\right]\)

= \(\left(\frac{10+8}{3}, \frac{-16+4}{3}, \frac{6-12}{3}\right)\)

= \(\left(\frac{18}{3}, \frac{-12}{3}, \frac{-6}{3}\right)\)

= (6, – 4, – 2).

Also, point S divides the join of PQ in the rano 2 : 1
∴ Coordinates of S are \(\left[\frac{2(10)+1(4)}{1+2}, \frac{2(-16)+1(2)}{1+2}, \frac{2(6)+1(-6)}{1+2}\right]\)

= \(\left(\frac{20+4}{3}, \frac{-32+2}{3}, \frac{12-6}{3}\right)\)

= \(\left(\frac{24}{3}, \frac{-30}{3}, \frac{6}{3}\right)\)

= (8, – 10, 2).

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 12 Introduction to Three Dimensional Geometry Ex 12.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry Ex 12.2

Question 1.
Find the distance between the following pairs of points :
(i) (2, 3, 5) and (4, 3, 1)
(ii) (- 3, 7, 2) and (2, 4, – 1)
(iii) (- 1, 3, – 4) and (1, – 3, 4)
(iv) (2, – 1, 3) and (- 2, 1, 3).
Answer.
The distance between point P(x₁, y₁, z₁) and Q(x₂, y₂, z₂) is given by
PQ = \(\)
(i) Distance between points (2, 3, 5) and (4, 3, 1)
= \(\sqrt{(4-2)^{2}+(3-3)^{2}+(1-5)^{2}}=\sqrt{(2)^{2}+(0)^{2}+(-4)^{2}}\)
= \(\sqrt{4+16}=\sqrt{20}=2 \sqrt{5}\)

(ii) Distance between points (- 3, 7, 2) and (2, 4, – 1)
= \(\sqrt{(2+3)^{2}+(4-7)^{2}+(-1-2)^{2}}=\sqrt{(5)^{2}+(-3)^{2}+(-3)^{2}}\)
= \(\sqrt{25+9+9}=\sqrt{43}\)

(iii) Distance between points ( – 1, 3, – 4) and (1, – 3, 4)
= \(\sqrt{(1+1)^{2}+(-3-3)^{2}+(4+4)^{2}}\)
= \(\sqrt{(2)^{2}+(-6)^{2}+(8)^{2}}\)
= \(\sqrt{4+36+64}=\sqrt{104}=2 \sqrt{26}\)

(iv) Distance between points (2, -1, 3) and (-2, 1, 3)
= \(\sqrt{(-2-2)^{2}+(1+1)^{2}+(3-3)^{2}}\)
= \(\sqrt{(-4)^{2}+(2)^{2}+(0)^{2}}\)
= \(\sqrt{16+4}=\sqrt{20}=2 \sqrt{5}\)

Question 2.
Show that the point (- 2, 3, 5), (1, 2, 3) and (7,0, – 1) are collinear.
Answer.
Let points (- 2, 3, 5), (1, 2, 3) and (7, 0, – 1) be denoted by P, Q, and R respectively.
Points P, Q, and R are collinear if they lie on a line.

Here, PQ + QP = \(\sqrt{14}+2 \sqrt{14}=3 \sqrt{14}\) = PR
Hence, points P (- 2, 3, 5), Q(1, 2, 3), and P (7, 0, – 1) are collinear.

Question 3.
Verify the following:
(i) (0, 7, – 10), (1, 6, – 6) and (4, 9, – 6) are the vertices of an isosceles triangle.
(ii) (0, 7, 10), (- 1, 6, 6) and (- 4, 9, 6) are the vertices of a right angled triangle.
(iii) (- 1, 2, 1), (1, – 2, 5), (4, – 7, 8) and (2, – 3, 4) are the vertices of a parallelogram.
Answer.
(i) Let points (0, 7, – 10), (1, 6, – 6) and (4, 9, – 6) be denoted by A, B and C respectively.

Here, AB = BC ≠ CA.
Thus, the given points are the vertices of an isosceles triangle.

(ii) Let (0, 7, 10), (- 1, 6, 6) and (- 4, 9, 6) be denoted by A, B, and C respectively.

Now, AB² + BC² = (3√2)² + (3√2)²
= 18 + 18 = 36 = AC²
Hence, the given points are the vertices of a right-angled triangle.

(iii) Let points (- 1, 2, 1), (1, – 2, 5), (4, – 7, 8) and (2, – 3, 4) be denoted by A, B, C and D respectively.

Now, AB = CD = 6, BC = AD = √43.
Hence, the opposite sides of quadrilateral ABCD, whose vertices are taken in order, are equal.
Therefore, ABCD is a parallelogram.
Hence, the given points are the vertices of a parallelogram.

Question 4.
Find the equation of the set of points which are equidistant from the points (1, 2, 3) and (3, 2, – 1).
Answer.
Let P(x, y, z) be the point that is equidistant from points A(l, 2, 3) and B(3, 2, -1).
Accordingly, PA = PB
⇒ PA² = PB²
⇒ (x – 1)² + (y – 2)² + (z – 3)² = (x – 3)² + (y – 2)² + (z + 1)²
x² – 2x + 1 + y² – 4y + 4 + z² – 6z + 9 = x² – 6x + 9 + y² – 4y + 4 + z² + 2z +1
⇒ – 2x – 4y – 6z + 14 = – 6x – 4y + 2z + 14
⇒ – 2x – 6z + 6x – 2z = 0
⇒ 4x – 8z = 0
⇒ x – 2z = 0
Thus, the required equation is x – 2z = 0.

Question 5.
Find the equation of the set of points P, the sum of whose distances from A (4, 0, 0) and B (- 4, 0, 0) is equal to 10.
Answer.
We have, a point P(x, y, z) such that, PA + PB = 0

On squaring both sides, we get
x² + y² + z² – 8x + 16 = 100 + x² + y² + z² + 8x + 16 – 20 \(\sqrt{x^{2}+y^{2}+z^{2}+8 x+16}\)
⇒ – 16x – 100 = – 20 \(\sqrt{x^{2}+y^{2}+z^{2}+8 x+16}\)
⇒ 4x + 25 = 5 \(\sqrt{x^{2}+y^{2}+z^{2}+8 x+16}\) [dividing both sides by -4]
Again squaring on both sides, we get
16x² + 200x + 625 = 25 (x² + y² + z² + 8x +16)
⇒ 16x² + 200x + 625 = 25x² + 25y² + 25z² +200x + 400
⇒ 9x² + 25y² + 25z² – 225 = 0.

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 12 Introduction to Three Dimensional Geometry Ex 12.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry Ex 12.1

Question 1.
A point is on the x-axis. What are its y – coordinates and z – coordinates?
Answer.
Coordinates of any point on the X-axis is (x, 0, 0).
Because at X-axis, both y and z – coordinates are zero.
So, its y and z – coordinates are zero.

Question 2.
A point is in the XZ – plane. What can you say about its y-coordinate?
Answer.
Any point on the XZ – plane will have the coordinate (x, 0, z), so its y – coordinate is 0.

Question 3.
Name the octants in which the following points lie : (1, 2, 3), (4, – 2, 3), (4, – 2, – 5), (4, 2, – 5), (- 4, 2, – 5), (- 4, 2, 5), (- 3, – 1, 6), (2, – 4, – 7).
Answer.
The x-coordinate, y-coordinate, and z-coordinate of point (1, 2, 3) are all positive. Therefore, this point lies in octant I.
The x-coordinate, y-coordinate, and z-coordinate of point (4, – 2, 3) are positive, negative, and positive respectively. Therefore, this point lies in octant IV.
The x-coordinate, y-coordinate, and z-coordinate of point (4, – 2, – 5) are positive, negative, and negative respectively. Therefore, this point lies in octant VIII.
The x-coordinate, y-coordinate, and z-coordinate of point (4, 2, – 5) are positive, positive, and negative respectively. Therefore, this point lies in octant V.
The x-coordinate, y-coordinate, and z-coordinate of point (- 4, 2, – 5) are negative, positive, and negative respectively. Therefore, this point lies in octant VI.
The x-coordinate, y-coordinate, and z-coordinate of point (- 4, 2, 5) are negative, positive, and positive respectively. Therefore, this point lies in octant II.
The x-coordinate, y-coordinate, and z-coordinate of point ( – 3, – 1, 6) are negative, negative, and positive respectively. Therefore, this point lies in octant III.
The x-coordinate, y-coordinate, and z-coordinate of point (2, – 4, – 7) are positive, negative, and negative respectively. Therefore, this point lies in octant VIII.

Question 4.
Fill in the blanks :
(i) The x-axis and y-axis taken together determine a plane known as
(ii) The coordinates of points in the XY-plane are of the form
(iii) Coordinate planes divide the space into octants.
Answer.
(i) XY plane
(ii) (x, y, 0)
(iii) eight

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 11 Conic Sections Miscellaneous Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 11 Conic Sections Miscellaneous Exercise

Question 1.
If a parabolic refletor is 20 cm in diameter and 5 cm deep, find the focus.
Answer.

Taking vertex of the parabolic reflector at origin, X-axis along the axis of parabola.
The equation of the parabola is of the fonn y² = 4ax.
Given, depth is 5 cm and diameter is 20 cm.
∵ Point P(5, 10) lies on parabola.
∴ (10)² = 4a(5) = a = 5
Clearly, focus is at the mid-point of given diameter i.e., S(5, 0).

Question 2.
An arch is in the form of a parabola with its Rxi vertical. The arch is 10 m high and 5 m wide at the base. How wide Is it 2m from the vertex of the parabola?
Answer.

Let the vertex of the parabola be at the origin and axis be along OY.
Then, the equation of the parabola is
x² = 4ay
The coordinates of end A of the (2.5, 10) and it lies on the eq. (i).
∴ (25)² = 4a × 10
⇒ a = \(\frac{6.25}{40}=\frac{5}{32}\) ……………(ii)

On putting the value of a from eq. (ii) in eq.(i), we get
x² = 4(\(\frac{5}{32}\))y ………………..(iii)

On substituting y = 2 in eq. (iii), we get
x² = \(\frac{5}{8}\) × 2
⇒ x² = \(\frac{5}{4}\)
⇒ x = \(\frac{\sqrt{5}}{2}\) m.
Hence, the width of the arc at a height of 2 m from vertex is 2 × \(\frac{\sqrt{5}}{2}\) i.e., √5 m.

Question 3.
The cable of a uniformly loaded suspension bridge hangs in the form of a parabola. The roadway which is horizontal and 100 m long is supported by vertical wires attached to the cable, the longest wire being 30 m and the shortest being 6m. Find the length of a supporting wire attached to the roadway 18 m from the middle.
Answer.

The cable is in the form of a parabola x² = 4ay.
Focus is at the middle of the cable, the shortest and longest vertical supports are 6 m and 30 m and roadway is 100 m long.
Since, point A(50, 24) lies on parabola x² = 4ay.
(50)² = 4a(24)
= 4a = 625
Equation of parabola is x² = \(\frac{625}{6}\) y
[Put 4a = \(\frac{625}{6}\)]
Let the support at 18 m from middle be l m, then B(18, l – 6) lies on the parabola.
∴ (18)² = \(\frac{625}{6}\) (l – 6)
l = \(\frac{18 \times 18 \times 6}{625}\) + 6
= 3.11 + 6 = 9.11 (approx)
Hence, the length of supporting wire is 9.11 m.

Question 4.
An arch is in the form of a semi-ellipse. It is 8 m wide and 2m high at the centre. Find the height of the arch at a point 1.5 m from one end.
Answer.

Clearly, equation of ellipse is of the form \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1 ……………(i)
Here, it is given that, 2a = 8 and b = 2.
⇒ a = 4 and b =2
On putting the values of a and b in eq. (i), we get
\(\frac{x^{2}}{16}+\frac{y^{2}}{4}\) = 1
which is required equation.
Given, AP = 1.5 m
∴ OP = OA – AP = 4 – 15
OP = 25m
Let PQ = k
∴ Coordinate of Q (2.5, k) will satisfy the equation of ellipse.

Question 5.
A rod of length 12 cm moves with its ends always touching the coordinate axes. Determine the equation of the locus of a point P on the rod, which is 3 cm from the end in contact with the x-axis.
Answer.

Let l be the length of the rod and at any position meet X-axis at A(a, 0) and Y-axis at B(0, b), so that
l² = a² + b²
⇒ (12)² = a² + b² …………….(i) [∵ l = 12]
Let P be the point on AB which is 3 cm from A and hence 9 cm from B.

This means that P divides AB in ratio 3 : 9 i.e., 1 : 3.
If P = (x, y), then by section formula, we have

(x, y) = \(\left(\frac{1 \times 0+3 \times a}{1+3}, \frac{1 \times b+3 \times 0}{1+3}\right)\)

(x, y) = \(\left(\frac{3 a}{4}, \frac{b}{4}\right)\)

⇒ x = \(\frac{3 a}{4}\)
⇒ a = \(\frac{4 x}{3}\) and b = 4y
On putting the values of a and b in eq. (i), we get
144 = (\(\frac{4 x}{3}\))² + (4 y)²
⇒ 1 = \(\frac{x^{2}}{9 \times 9}+\frac{y^{2}}{9}\)

⇒ \(\frac{x^{2}}{81}+\frac{y^{2}}{9}\) = 1
which is required equation.

Question 6.
Find the area of the triangle formed by the lines joining the vertex of the parabola x² = 12y to the ends of its latus rectum.
Answer.
The given parabola is x² = 12y.
On comparing this equation with x² = 4ay, we obtain
4a = 12
⇒ a = 3.
The coordinates of foci are S (0, a) = S (0, 3)

Let AB be the latus rectum of the given parabola.
The given parabola can be roughly drawn as
At y = 3, x² = 12 (3)
x² = 36
⇒ x = ± 6.
The coordinates of A are (- 6, 3), while the coordinates of B are (6, 3).
Therefore, the vertices of ∆OAB are 0 (0, 0), A (- 6, 3), and B (6, 3).
Area of ∆OAB = \(\frac{1}{2}\) |0 (3 – 3) + (- 6) (3 – 0) + 6 (0 – 3)|
= \(\frac{1}{2}\) |(- 6)(3) + 6(- 3)| unit²
= \(\frac{1}{2}\) |- 18 – 18| unit²
= \(\frac{1}{2}\) |- 36| unit²
= \(\frac{1}{2}\) × 36 unit²
= 18 unit²
Thus, the required area of the triangle is 18 unit².

Question 7.
A man running a racecourse notes that the sum of the distances from the two flag posts from him is always 10 m and the distance between the flag posts is 8 m. Find the equation of the posts traced by the man.
Answer.

Let A and B be the positions of the two flag posts and P(x, y) be the position of the man. Accordingly, PA + PB = 10.
We know that if a point moves in a plane in such a way that the sum of its distances from two fixed points is constant, then the path is an ellipse and this constant value is equal to the length of the major axis of the ellipse.

Therefore, the path described by the man is an ellipse where the length of the major axis is 10 m, while points A and B are the foci.
Taking the origin of the coordinate plane as the centre of the ellipse, while taking the major axis along the x-axis, the ellipse can be diagrammatically represented as.
The equation of the ellipse will be of the form \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1 , where a is the semi-major axis.
Accordingly, 2a = 10
⇒ a = 5.
Distance between the foci, 2c = 8
⇒ c = 4
On using the relation c = \(\sqrt{a^{2}-b^{2}}\) we obtain 4 = \(\sqrt{25-b^{2}}\)
⇒ 16 = 25 – b²
⇒ b = 25 – 16 = 9
⇒ b = 3
Thus, the equation of the path traced by the man is \(\frac{x^{2}}{25}+\frac{y^{2}}{9}\) = 1.

Question 8.
An equilateral triangle is inscribed in the parabola y =4ax, where one vertex is at the vertex of the parabola. Find the length of the side of the triangle.
Answer.
Given, equation of parabola is y² = 4ax.
Let p be the side of equilateral ∆OAB, whose one vertex is the vertex of parabola.
Then, by symmetry, AB is perpendicular to the axis ON of parabola.
Let ON = x, then BN = \(\frac{p}{2}\)
Since, B(x, \(\frac{p}{2}\)) lies on parabola.

⇒ p² = 64 × 3 × a²
⇒ p = 8a√3
Hence, side of an equilateral triangle is 8a√3.

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 11 Conic Sections Ex 11.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.4

Question 1.
Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola
\(\frac{x^{2}}{16}-\frac{y^{2}}{9}\) = 1.
Answer.
The given equation is \(\frac{x^{2}}{16}-\frac{y^{2}}{9}\) = 1 or \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1.

On comparing this equation with the standard equation of hyperbola i.e., \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1, we obtain a = 4 and b = 3.

We know that
a² + b² = c²
4² + 3² = 25
⇒ c = 5
Therefore, The coordinates of the foci are (± 5, 0).
The coordinates of the vertices are (± 4, 0).
Eccentricity e = \(\frac{c}{a}=\frac{5}{4}\)

Length of latusrectum = \(\frac{2 b^{2}}{a}=\frac{2 \times 9}{4}=\frac{9}{2}\).

Question 2.
Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola
\(\frac{y^{2}}{9}-\frac{x^{2}}{27}\) = 1.
Answer.
The given equation is \(\frac{y^{2}}{9}-\frac{x^{2}}{27}\) = 1 or \(\frac{y^{2}}{3^{2}}-\frac{x^{2}}{(\sqrt{27})^{2}}\) = 1.
On comparing this equation with the standard equation of hyperbola i.e., \(\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}\) = 1, we obtain a = 3 and b = √27.

We know that a² + b² = c²
c² = 3² + (√27)²
= 9 + 27 = 36
⇒ c = 6
Therefore, the coordinates of the foci are (0, ± 6).
The coordinates of the vertices are (0, ± 3).
Eccentricity, e = \(\frac{c}{a}=\frac{6}{3}\) = 2.
Length of latusrectum = \(\frac{2 b^{2}}{a}=\frac{2 \times 27}{3}\) = 18.

Question 3.
Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola
9y² – 4x² = 36.
Answer.
The given equation is 9y² – 4x² = 36
It can be written as 9y² – 4x² = 36
or \(\frac{y^{2}}{4}-\frac{x^{2}}{9}\) = 1

or \(\frac{y^{2}}{2^{2}}-\frac{x^{2}}{3^{2}}\) = 1 ………….(i)
On comparing equation (i) with the standard equation of hyperbola i.e., \(\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}\) = 1, we obtain a = 2 and b = 3.

We know that a² + b² = c²
c² = 4 + 9 = 13
⇒ c = √13.
Therefore, the coordinates of the foci are (0, ± √13).
The coordinates of the vertices are (0, ± 2).
Eccentricity, e = \(\frac{c}{a}=\frac{\sqrt{13}}{2}\)

Length of latus rectum = \(\frac{2 b^{2}}{a}=\frac{2 \times 9}{2}\) = 9.

Question 4.
Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola 16x² – 9y² = 576.
Answer.
The given equation is 16x² – 9y² = 576
It can be written as 16x² – 9y² = 576

⇒ \(\frac{x^{2}}{36}-\frac{y^{2}}{64}\) = 1

⇒ \(\frac{x^{2}}{6^{2}}-\frac{y^{2}}{8^{2}}\) = 1 ………………(i)

On comparing equation (i) with the standard equation of hyperbola i.e.,
\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1, we obtain a = 6 and b = 8.
We know that a² + b² = c²
c² = 36 + 64 = 100
⇒ c = 10
Therefore, the coordinates of the foci are (± 10, 0).
The coordinates of the vertices are (± 6, 0).
Eccentricity, e = \(\frac{c}{a}=\frac{10}{6}=\frac{5}{3}\).

Length of larus rectum = \(\frac{2 b^{2}}{a}=\frac{2 \times 64}{6}=\frac{64}{3}\).

Question 5.
Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola 5y² – 9x² = 36.
Answer.
The given equation is 5y² – 9x² = 36.
⇒ \(\frac{y^{2}}{\left(\frac{36}{5}\right)}-\frac{x^{2}}{4}\) = 1

⇒ \(\frac{y^{2}}{\left(\frac{6}{\sqrt{5}}\right)^{2}}-\frac{x^{2}}{2^{2}}\) = 1
On comparing equation (i) with the standard equation of hyperbola i.e., \(\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}\) = 1,

we obtain a = \(\frac{6}{\sqrt{5}}\) = and b = 2.

We know that a² + b² = c².
∴ c² = \(\frac{36}{5}+4=\frac{56}{5}\)

⇒ c = \(\sqrt{\frac{56}{5}}=\frac{2 \sqrt{14}}{\sqrt{5}}\)

Therefore, the coordinates of the foci are (0, ± \(\frac{2 \sqrt{14}}{\sqrt{5}}\))

The coordinates of the vertices are (o, ± \(\frac{6}{\sqrt{5}}\)).

Eccentricity, e = \(\frac{c}{a}=\frac{\left(\frac{2 \sqrt{14}}{\sqrt{5}}\right)}{\left(\frac{6}{\sqrt{5}}\right)}=\frac{\sqrt{14}}{3}\)

Length of latus rectum = \(\frac{2 b^{2}}{a}=\frac{2 \times 4}{\left(\frac{6}{\sqrt{5}}\right)}=\frac{4 \sqrt{5}}{3}\)

Question 6.
Find the coordinates of the foci, and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola
49y² – 16x² = 784.
Answer.
The given equation is 49y² – 16x² = 784
it can be writtern as 49y² – 16x² = 784

or \(\frac{y^{2}}{16}-\frac{x^{2}}{49}\) = 1

or \(\frac{y^{2}}{4^{2}}-\frac{x^{2}}{7^{2}}\) = 1 ……………….(i)

On comparing equation (i) with the standard equation of hyperbola i.e., \(\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}\) = 1,
we obtain a = 4 and b = 7.

We knowthat, a² + b² = c².
∴ c² = 16 + 49 = 65
⇒ c = √65
Therefore, The coordinates of the foci are (0, ± √65
The coordinates of the vertices are (0, ± 4).

Eccentricity, e = \(\frac{c}{a}=\frac{\sqrt{65}}{4}\).

Length of lattis rectum = \(\frac{2 b^{2}}{a}=\frac{2 \times 49}{4}=\frac{49}{2}\).

Question 7.
Find the equation of the hyperbola satisfying the given conditions : Vertices (± 2, 0), foci (± 3, 0).
Answer.
Given, vertices (± 2, 0), foci (± 3, 0)
Here, the vertices are on the x-axis.
Therefore, the equation of the hyperbola is of the form \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1.
Since the vertices are (± 2, 0), a = 2
Since the foci are (± 3, 0), c = 3
We know that a² + b² = c².
∴ 2² + b² = 3²
b² = 9 – 4 = 5
Thus, the equation of the hyperbola is \(\frac{x^{2}}{4}-\frac{y^{2}}{5}\)= 1.

Question 8.
Find the equation of the hyperbola satisfying the given conditions : Vertices (0, ± 5), foci (0, ± 8).
Ans.
We have,vertices (0, ± a) = (0, ± 5) = a = 5
and foci(0, ± c) = (0, ± 8)
⇒ c = 8
We know that, c² = a² + b²
⇒ 64 = 25 + b²
⇒ b² = 64 – 25
⇒ b² = 39
Here, the foci and vertices lie on Y-axis, therefore the equation of hyperbola is of the form \(\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}\) = 1
i.e. \(\frac{y^{2}}{25}-\frac{x^{2}}{39}\) = 1

Question 9.
Find the equation of the hyperbola satisfying the given conditions : Vertices (0, ± 3), foci (0, ± 5).
Answer.
Given, vertices (0, ± 3), foci (0, ± 5)
Here, the vertices are on the y-axis.
Therefore, the equation of the hyperbola is of the form \(\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}\) = 1.
Since the vertices are (0, ± 3), a = 3.
Since the foci are (0, ± 5), c = 5.
We know that a² + b² = c².
3² + b² = 5²
⇒ b² = 25 – 9 = 16.
Thus, the equation of the hyperbola is \(\frac{y^{2}}{9}-\frac{x^{2}}{16}\) = 1.

Question 10.
Find the equation of the hyperbola satisfying the given conditions : Foci (± 5, 0), the transverse axis is of length 8.
Answer.
Given, foci (± 5, 0), the transverse axis is of length 8.
Here, the foci are on the x-axis.
Therefore, the equation of the hyperbola is of the \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1.
Since the foci are (± 5, 0), c = 5.
Since the length of the transverse axis is 8, 2a = 8
⇒ a = 4.
We know that, a² + b² = c².
∴ 4² + b² = 5²
⇒ b² = 25 – 16 = 9

Question 11.
Find the equation of the hyperbola satisfying the given conditions : Foci (0, ± 13), the conjugate axis is of length 24. Answer.
Given, foci (0, ± 13), the conjugate axis is of length 24.
Here, the foci are on the y-axis.
Therefore, the equation of the hyperbola is of the form \(\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}\) = 1.
a b2
Since the foci are (0, ± 13), c = 13.
Since the length of the conjugate axis is 24, 2b = 24
⇒ b = 12.
We know that a² + b² = c².
a² + 12² = 13²
⇒ a² = 169 – 144 = 25
Thus, the equation of the hyperbola is \( \frac{y^{2}}{25}-\frac{x^{2}}{144}\) = 1.

Question 12.
Find the equation of the hyperbola satisfying the given conditions : Foci (± 3√5, 0), the latus rectum is of length 8.
Answer.

Here, foci at (± 3√5, 0)
∴ (± c, 0) = (± 3√5, 0)
⇒ c = 3√5
Length of latus rectum = \(\frac{2 b^{2}}{a}\) = 8 [given]
∴ b² = 4a …………….(i)
We know that, c² = a² + b²
⇒ (3√5)² = a² + 4a [put c = 3√5]
⇒ a = – 9, a = 5
∴ a = 5 as a cannot be negative.
On putting a = 5 in eq. (i), we get
b² = 5 × 4
⇒ b² = 20
Since, foci lies on X-axis, therefore the equation of hyperbola is of the form
\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1

i.e., \(\frac{x^{2}}{25}-\frac{y^{2}}{20}\) = 1 [Put a² = 5 and b² = 20]
⇒ 20x² – 25y² = 500
⇒ 4x² – 5y² = 100 [dividing both sides by 5]
which is required equation of hyperbola.

Question 13.
Find the equation of the hyperbola satisfying the given conditions : Foci (± 4, 0), the latus rectum is of length 12.
Answer.
Given, foci (± 4, 0), the latus rectum axis is of length 12.
Here, the foci are on the x-axis.
Therefore, the equation of the hyperbola is of the form \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1.
Since the foci are (± 4, 0), c = 4.
Length of latus rectum =12
⇒ \(\frac{2 b^{2}}{a}\) = 12
⇒ b² = 6a
We know that a² + b² = c².
a² + 6a = 16
⇒ a² + 6a – 16 = 0
⇒ a² + 8a – 2a – 16 = 0
⇒ (a + 8) (a – 2) = 0
⇒ a = – 8, 2
Since a is non-negative, a = 2.
b² = 6a = 6 × 2 = 12
Thus, the equation of the hyperbola is \(\frac{x^{2}}{4}-\frac{y^{2}}{12}\) = 1.

Question 14.
Find the equation of the hyperbola satisfying the given conditions : Vertices (± 7, 0), e = \(\frac{4}{3}\).
Answer.
Given, vertices (± 7, 0), e = \(\frac{4}{3}\).
Here, the foci are on the x-axis.
Therefore, the equation of the hyperbola is of the form \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1.
Since the vertices are (± 7, 0), a = 7.

Question 15.
Find the equation of the hyperbola satisfying the given conditions : Foci (0, ± √10), passing through (2, 3).
Answer.
Since, the foci (0, ± √10) are along Y-axis, then equation of hyperbola is of the form

9(10 – a²) – 4a² = a²(10 – a²)
⇒ 90 – 13a² = 10a² – a⁴
⇒ a⁴ – 23a² + 90 = a
⇒ (a² – 18) (a² – 5) = 0
∴ a² = 18 or 5
Since, e = \(\frac{\sqrt{10}}{a}\) and e > 1
∴ \(\frac{\sqrt{10}}{a}\) > 1
⇒ a < √10; a² < 10
we reject a² = 18
Thus, a² = 5
∴ b² = a² e² – a²
= 10 – 5 = 5
Hence, equation of hyperbola is \(\frac{y^{2}}{5}-\frac{x^{2}}{5}\) = 1 [put a² = b² = 5]
or y² – x² = 5.

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 11 Conic Sections Ex 11.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.2

Question 1.
Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length .of the latus rectum for y² = 12x.
Answer.
The given equation is y² = 12x.
Here, the coefficient of x is positive.
Hence, the parabola opens towards the right.
On comparing this equation with y² = 4 ax, we obtain
4a = 12
⇒ a = 3.
∴ Coordinates of the focus = (a, 0) = (3, 0)
Since the given equation involves y², the axis of the parabola is the x-axis.
Equation of directrix, x = – a i.e., x = – 3 i.e., x + 3 = 0
Length of latus rectum = 4a = 4 × 3 = 12.

Question 2.
Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum for x² = 6y.
Answer.
The given equation is x² = 6y.
Here, the coefficient of y is positive. Hence, the parabola opens upwards.
On comparing this equation with x² = 4ay , we obtain
4a = 6
⇒ a = \(\frac{3}{2}\).
Coordinates of the focus = (0, a) = (0, \(\frac{3}{2}\))
Since the given equation involves x , the axis of the parabola is the y-axis.
Equation of directrix, y = – a i.e., y = – \(\frac{3}{2}\)
Length of latus rectum = 4a = 6.

Question 3.
Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum for y² = – 8x.
Answer.
Here, the coefficient of x is negative.
Hence, the parabola opens towards the left.
On comparing this equation with y² = – 4 ax , we obtain
– 4a = – 8
⇒ a = 2
Coordinates of the focus = (- a, 0) = (- 2, 0)
Since the given equation involves y², the axis of the parabola is the x-axis.
Equation of directrix, x = a i.e., x = 2
∴ Length of latus rectum = 4a = 8.

Question 4.
Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum for x² = – 16y.
Ans.
The given equation is x² = – 16y
Here, the coefficient of y is negative.
Hence, the parabola opens downwards.
On comparing this equation with x² = – 4ay, we obtain
– 4a = – 16
⇒ a = 4
∴ Coordinates of the focus = (0, – a) = (0, – 4)
Since the given equation involves x², the axis of the parabola is the y-axis.
Equation cf directrix, y = a i.e., y = 4
Length of latus rectum = 4a = 16.

Question 5.
Find the coordinntes of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum for y² = 10x.
Answer.
The given equation is y² = 10x.
Here, the coefficient of x is positive.
Hence, the parabola opens towards the right.
On comparing this equation with y² = 4ax, we obtain
4a = 10
⇒ a = \(\frac{5}{2}\)
Coordinates of the focus = (a, 0) = (\(\frac{5}{2}\), 0)
Since the given equation involves y², the axis of the parabola is the x-axis.
Equation of directrix, x = – a, i.e., x = – \(\frac{5}{2}\)
Length of latus rectum = 4a = 10.

Question 6.
Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum for x\(\frac{5}{2}\) = – 9y.
Answer.
We have, equation of parabola is x² = – 9y
So, axis of symmetry is along x-axis.
The coefficient of y is negative, so parabola opens downward.
On comparing with the equation,
x² = – 4ay, we get a = \(\frac{9}{4}\)
Thus, the focus is (0, – a) i.e., (0, – \(\frac{9}{4}\))
Equation of directrix is y = a i.e., y = 9/4.
Length of latus rectum = 4a i.e., 9.

Question 7.
Find the equation of the parabola that satisfies the given conditions : Focus (6, 0): directrix x = – 6.
Answer.
Given, focus (6, 0) and directrix, x = – 6.
Here, we see that in focus, x-coordinate is positive and y-coordinate is zero.
So, focus lies on the positive direction of X-axis i.e., equation of parabola will be of the form y² = 4ax with a = 6.
Hence, required equation is
y² = 4 × 6x
⇒ y² = 24x

Question 8.
Find the equation of the parabola that satisfies the given conditions : Focus (0, – 3); directrix y = 3.
Answer.
Given, focus = (0, – 3); directrix y=3.
Since the focus lies on the y-axis, the y-axis is the axis of the parabola.
Therefore, the equation of the parabola is either of the form x² = 4 ay or x² = – 4ay.
It is also seen that the directrix, y = 3 is above the x-axis, while the focus (0, – 3) is below the x-axis.
Hence, the parabola is of the form x²= – 4ay.
Here, a = 3.
Thus, the equation of the parabola is x² = – 12y.

Question 9.
Find the equation of the parabola that satisfies the given conditions : Vertex (0, 0); focus (3, 0).
Answer.
Given, vertex (0, 0); focus (3, 0).
Since the vertex of the parabola is (0, 0) and the focus lies on the positive x-axis, x-axis is the axis of the parabola, while the equation of the parabola is of the form y² = 4ax.
Since the focus is (3, 0), a = 3.
Thus, the equation of the parabola is y² = 4 × 3 × x, i.e., y² = 12x.

Question 10.
Find the equation of the parabola that satisfies the given conditions: Vertex (0, 0) focus (-2, 0).
Answer.
Given, vertex (0, 0); focus (- 2, 0)
Since the vertex of the parabola is (0, 0) and the focus lies on the negative x-axis, x-axis is the axis of the parabola, while the equation of the parabola is of the form y² = – 4ax.
Since the focus is (- 2, 0), a = 2.
Thus, the equation of the parabola is y² = – 4(2)x, i.e., y = – 8x.

Question 11.
Find the equation of the parabola that satisfies the given conditions: Vertex (0, 0) passing through (2, 3) and axis is along x-axis.
Answer.
Since the vertex is (0, 0) and the axis of the parabola is the x-axis, the equation of the parabola is either of the form
y² = 4ax or y² = – 4ax.
The parabola passes through point (2, 3), which lies in the first quadrant.
Therefore, the equation of the parabola is of the form y² = 4 ax, while point (2, 3) must satisfy the equation y² = 4ax.
3² = 4a(2)
a = \(\frac{9}{8}\)
Thus, the equation of the paraola is y² = – 4(2)x, i.e., y² = – 8x.

Question 12.
Find the equation of the parabola that statisfies the given conditions; Vertex (0, 0), passing through (5, 2) and symmetric with respect to y-axis.
Answer.

Since, the parabola is symmetric about Y-axis and has its vertex at the origin, so the equation is of the form x² = 4ay or x² = – 4ay.
But parabola passes through (5, 2) which lies in the first quadrant. It must open upward.
Thus, the equation is of the form
x² = 4ay ……………….(i)
Since, parabola passes through (5, 2)
(5)² = 4a(2)
⇒ 25 = 8a
⇒ a = \(\frac{25}{8}\)
On putting the value of a in eq.(i), we get
x² = 4 (\(\frac{25}{8}\)) y
⇒ x² = \(\frac{25}{8}\) y

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 11 Conic Sections Ex 11.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.1

Question 1.
Find the equation of the circle with centre (0, 2) and radius 2.
Answer.
The equation of a circle with centre (h, k) and radius r is given as
(x – h)² + (y – k)² = r²
It is given that centre (h, k) = (0, 2) and radius (r) = 2
Therefore, the equation of the circle is
(x – 0)² + (y – 2)² = 2²
x² + y² + 4 – 4y = 4
x² + y² – 4y = 0.

Question 2.
Find the equation of the circle with centre (- 2, 3) and radius 4.
Answer.
The equation of a circle with centre (h, k) and radius r is given as
(x – h)² + (y – k)² = r²
It is given that centre (h, k) = (- 2, 3) and radius (r) = 4.
Therefore, the equation of circle is
(x + 2)² + (y – 3)² = (4)²
x² + 4x + 4 + y² – 6y + 9 = 16
x² + y² + 4x – 6y – 3 = 0.

Question 3.
Find the equation of the circle with centre (\(\frac{1}{2}\), \(\frac{1}{4}\)) and radius \(\frac{1}{12}\).
Answer.

Question 4.
Find the equation of the circle with cehtre (1,1) and radius √2.
Answer.

Here, centre is at (1, 1).
∴ h = 1, k = 1
Also, radius, r = √2 units
[∵ equation of circle having centre (h, k) and radius r is (x – h)² + (y – k)² = r²].
Equation of circle is (x – 1)² + (y – 1)² = (V2)²
⇒ x² – 2x + 1 + y² – 2y + 1 = 2
⇒ x² + y² – 2x – 2y = 0

Question 5.
Find the equation of the circle with centre (- a, – b) and radius \(\sqrt{a^{2}-b^{2}}\).
Answer.
The equation of a circle with centre (h, k) and radius r is given as (x – h)² + (y – k)² = r²
It is given that centre (h, k) = (- a, – b) and radius (r) = \(\sqrt{a^{2}-b^{2}}\).
Therefore, the equation of the circle is (x + a)² + (y + b)² = (\(\sqrt{a^{2}-b^{2}}\))²
x² + 2ax + a² + y² + 2by + b² = a² – b²
x² + y² + 2ax + 2by + 2b² = 0.

Question 6.
Find the centre and radius of the circle (x + 5)² + (y – 3)² = 36.
Answer.
The equation of given circle is (x + 5)² + (y – 3)² = 36
(x + 5)² + (y – 3)² = 36
⇒ {x – (- 5)}² + (y – 3)² = (6)²,
which is of the form (x – h)² + (y – k)² = r²,
where h = – 5, k = 3, and r = 6.
Thus, the centre of the given circle is (- 5, 3), while its radius is 6.

Question 7.
Find the equation of the circle x² + y² – 4x – 8y – 45 = 0.
Answer.
The equation of given circle is x² + y² – 4x – 8y – 45 = 0.
x² + y² – 4x – 8y – 45 = 0
⇒ (x² – 4x) + (y² – 8y) = 45
⇒ {x² – 2 (x) (2) + 22} + {y² – 2(y) (4) + 42} – 4 – 16 = 45
⇒ (x – 2)² + (y – 4)² = 65 ,
⇒ (x – 2)² + (y – 4)² = (√65)², which is of the form
(x – h)² + (y – k)² = r² , where h = 2, k = 4 and r = √65
Thus, the centre of the given circle is (2, 4), while its radius is √65.

Question 8.
Find the centre and radius of the circle x² + y² – 8x + 10y – 12 = 0.
Answer.
The equation of the given circle is x² + y² + 8x + 10y – 12 = 0.
x² + y² – 8x + 10y – 12 = 0
(x² – 8x) + (y² + 10y) = 12
⇒ (x² – 2 (x) (4) + 4²) + (y² + 2(y) (5) + 5²) – 16 – 25 = 12
(x – 4)² + (y + 5)² = 53
(x – 4)² + {y – (- 5)}² = (√53)², which is of the form
(x – h)² + (y – k)² = r², where h = 4, k = – 5 and r = √53
Thus, the centre of the given circle is (2, 4), while its radius is √53.

Question 9.
Find the centre and radius of the circle 2x² + 2y² – x = 0.
Answer.
The given equation of circle is
2x² + 2y² – x = 0
x² + y² – \(\frac{x}{2}\) = 0
(x² – \(\frac{x}{2}\)) + y² = 0
On adding \(\frac{1}{16}\) to make perfect squares, we get
(x² – \(\frac{x}{2}\) + \(\frac{1}{16}\)) + y² = \(\frac{1}{16}\)
(x – \(\frac{1}{4}\))² + (y – 0)² = (\(\frac{1}{4}\))²
On comparing with (x – h)² + (y – k)² = r² , we get
h = \(\frac{1}{4}\), k = 0 and r = \(\frac{1}{2}\)
Centre = (h, k) = (\(\frac{1}{4}\), 0)
Radius = \(\frac{1}{4}\)

Question 10.
Find the equation of the circle passing through the points (4, 1) and (6, 5) and whose centre is on the line 4x + y = 16.
Answer.
Let the equation of the required circle be (x – h)² + (y – k)² = r².
Since the circle passes through points (4, 1) and (6, 5).
(4 – h)² + (1 – k)² = r² ……………(i)
(6 – h)² + (5 – k)² = r² ……………(ii)
Since the centre (h, k) of the circle lies on line 4x + y = 16,
4h + k = 16
From equations (i) and (ii), we obtain
(4 – h)² + (1 – k)² = (6 – h)² + (5 – k)²
⇒ 16 – 8h + h² + 1 – 2k + k² = 36 – 12h + h² + 25 – 10k + k²
⇒ 16 – 8h + 1 – 2k = 36 – 12h + 25 – 10k
⇒ 4h + 8k = 44
⇒ h + 2k = 11 …………………(iv)
On solving equations (iii) and (iv), we obtain h = 3 and k = 4.
On substituting the values of h and k in equation (i), we obtain
(4 – 3)² + (1 – 4)² = r²
⇒ (1)² + (- 3)² = r²
⇒ 1 + 9 = r²
⇒ r² = 10
⇒ r = √10
Thus, the equation of the required circle is (x – 3)² + (y – 4)²= (√10)²
x² – 6x + 9 + y² – 8y + 16 = 10
x² + y² – 6x – 8y + 15 = 0.

Question 11.
Find the equation of the circle passing through the points (2, 3) and (- 1, 1) and whose centre is on the line x – 3y – 11 = 0.
Answer.

Let the equation of the circle be
(x -h)² + (y – k)² =r² ……………(i)
Since, the circle (i) passes through (2, 3) and (- 1, 1).
We have, (2 – h)² + (3 – k)² = r²
⇒ h² – 4h + 4 + 9 – 6k + k² = r² and
(- 1 – h)² + (1 – k)² = r²
h² + 2h + 1 + 1 – 2k + k² = r²
Centre (h, k) of circle is on the line
x – 3y – 11 = 0
∴ h – 3k – 11 = 0 …………..(iv)
On subtracting eQuestion (iii) from eq.(ii), we get
– 6h – 4k + 11 = 0
⇒ 6h + 4k = 11 ……………….(v)
On solving eqs. (iv) and (v), we get
h = \(\frac{7}{2}\), k = \(\frac{-5}{2}\)
On putting the values of h and k in eq. (ii), we get
(2 – \(\frac{7}{2}\))² + (3 + \(\frac{5}{2}\))² = r²

(- \(\frac{7}{2}\))² + (\(\frac{11}{2}\))² = r²

Therefore, equation of circle having centre (\(\frac{7}{2}\), – \(\frac{5}{2}\)) and radius \(\frac{65}{2}\) is

(x – \(\frac{7}{2}\))² + (y + \(\frac{5}{2}\))² = \(\frac{65}{2}\)

x² + \(\frac{49}{4}\) – 7x + y² + \(\frac{25}{4}\) + 5y = \(\frac{65}{2}\)

x² + y² – 7x + 5y – \(\frac{56}{4}\) = 0
x² + y² – 7x + 5y – 14 = 0

Alternative Method:

Let the equation of circle be
x² + y² + 2gx + 2fy + c = 0 ……………..(i)
Since, it passes through (2, 3).
∴ (2)² + (3)² + 2g(2) + 2f(3) + c = 0
4 + 9 + 4g + 6f + c = 0
4g + 6f + c = – 13 ……………….(ii)
Circle also passes through (- 1, 1).
∴ (- 1)² + (1)² + 2g(- 1) + 2f(1) + c = 0
⇒ 1 + 1 – 2g + 2f + c = 0
⇒ – 2g + 2f + c = – 2 …………..(iii)
Since, centre of circle lies on line x – 3y – 11 = 0
i.e.,C(- g, – f) lies on line.
∴ – g – 3 (- f) – 11 = 0
– g + 3f = 11 …………..(iv)
On subtracting eq. (iii) from eq. (ii), we get
6g + 4f = – 11 …………….(v)
On solving eqs. (iv) and (y) for g and f, we get
g = – \(\frac{7}{2}\), f = \(\frac{5}{2}\)
On putting the values of g and f in eq. (ii), we get
⇒ – 14 + 15 + c = – 13
⇒ c = – 14
Now, putting the values of g, f and c in eq. (i), we get
x² + y² + 2(- \(\frac{7}{2}\)) + 2 (\(\frac{7}{2}\)) – 14 = 0
x² + y² – 7x + 5y – 14 = 0.

Question 12.
Find the equation of the circle with radius 5 whose centre lies on x-axis and passes through the point (2, 3).
Answer.
Let the equation of the required circle be (x – h)² + (y – k)² = r².
Since the radius of the circle is 5 and its centre lies on the x-axis, k = 0 and r = 5.
Now, the equation of the circle becomes (x – h)² + y² = 25.
It is given that the circle passes through point (2, 3).
(2 – h)² + 3² = 25
⇒ (2 – h)² = 25 – 9
⇒ (2 – h)² = 16
⇒ 2 – h = ± √l6 = ± 4
If 2 – h = 4, then h = – 2
If 2 – h = – 4, then h = 6
when h = – 2, the equation of the circle becomes
(x + 2)² + y² = 25
x² + 4x + 4 + y² = 25
x² + y² + 4x – 21 = 0.
When h = 6, the equation of the circle becomes
(x – 6)² + y² = 25
x² – 12x + 36 + y² = 25
x² + y² – 12x + 11 = 0.

Question 13.
Find the equation of the circle passing through (0, 0) and making intercepts a and b on the coordinate axes.
Answer.
Let the equation of the required circle be (x- h)² + (y – k)² = r².
Since the circle passes through (0, 0).
(0 – h)² + (0 – k)² = r².
h² + k² = r²
The equation of the circle now becomes (x – h)² + (y – k)² = h² + k².
It is given that the circle makes intercepts a and b on the coordinate axes.
This means that the circle passes thróugh points (a, 0) and (0, b).
(a – h)² + (0 – k)² = h² + k²
(0 – h)² + (b – k)² = h² + k²
From equation (i), we obtain
a² – 2ah + h² + k² = h² + k²
⇒ a² – 2ah = 0
⇒ a(a – 2h) = 0
⇒ a = 0 or (a – 2h) = 0
However, a ≠ 0; hence, (a – 2 h) = 0
⇒ h = \(\frac{a}{2}\).
From equation (ii), we obtain
h² + b² – 2bk + k² = h² + k²
⇒ b² – 2bk = 0
⇒ b(b – 2k) = 0
⇒ b = 0 or (b – 2k) = 0
However, b ≠ 0; hence, (b – 2k) = 0
⇒ k = \(\frac{b}{2}\).
Thus, the equation of the required circle is .
\(\left(x-\frac{a}{2}\right)^{2}+\left(y-\frac{b}{2}\right)^{2}=\left(\frac{a}{2}\right)^{2}+\left(\frac{b}{2}\right)^{2}\)

⇒ \(\left(\frac{2 x-a}{2}\right)^{2}+\left(\frac{2 y-b}{2}\right)^{2}=\frac{a^{2}+b^{2}}{4}\)

⇒ 4x² – 4ax + a² + 4y² – 4by + b² = a² + b²
⇒ 4x² + 4y² – 4ax – 4by = 0
⇒ x² + y² – ax – by = 0.

Question 14.
Find the equation of a circle with centre (2, 2) and passes through the point (4, 5).
Answer.
The equation of circle is (x – h)² + (y – k)² = r² ……………..(i)
Since the circle passes through point (4, 5) and co-ordinates of centre are (2, 2)
∴ radius of circle = \(\sqrt{(4-2)^{2}+(5-2)^{2}}=\sqrt{4+9}=\sqrt{13}\)
Now the equation of required circle is
(x – 2)² + (y – 2)² = (√13)²
4x² + 4y² – 4ax – 4ay = 13
x² + y² – 4x – 4y – 5 = 0

Question 15.
Does the point (- 2.5, 3.5) lie inside, outside or on the circle x² + y² = 25?
Answer.
The equation of the given circle is x² + y² = 25
x² + y² = 25
⇒ (x – 0)² + (y – 0)² = 5², which is of the form(x – h)² + (y – k)² = r²,
where h = 0, k = 0 and r = 5.
centre = (0, 0) and radius = 5.
Distance between point (- 2.5, 3.5) and centre (0, 0)
= \(\sqrt{(-2.5-0)^{2}+(3.5-0)^{2}}=\sqrt{6.25+12.25}=\sqrt{18.5}\)
= 4.3 (approx.) < 5.
Since the distance between point (- 2.5, 3.5) and centre (0, 0) of the circle is less than the radius of the circle, point (- 2.5, 3.5) lies inside the circle.