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

a^{2} + b^{2} = c^{2}

4^{2} + 3^{2} = 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 a^{2} + b^{2} = c^{2}

c^{2} = 3^{2} + (√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

9y^{2} – 4x^{2} = 36.

Answer.

The given equation is 9y^{2} – 4x^{2} = 36

It can be written as 9y^{2} – 4x^{2} = 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 a^{2} + b^{2} = c^{2}

c^{2} = 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 16x^{2} – 9y^{2} = 576.

Answer.

The given equation is 16x^{2} – 9y^{2} = 576

It can be written as 16x^{2} – 9y^{2} = 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 a^{2} + b^{2} = c^{2}

c^{2} = 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 5y^{2} – 9x^{2} = 36.

Answer.

The given equation is 5y^{2} – 9x^{2} = 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 a^{2} + b^{2} = c^{2}.

∴ c^{2} = \(\frac{36}{5}+4=\frac{56}{5}\)

⇒ 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

49y^{2} – 16x^{2} = 784.

Answer.

The given equation is 49y^{2} – 16x^{2} = 784

it can be writtern as 49y^{2} – 16x^{2} = 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, a^{2} + b^{2} = c^{2}.

∴ c^{2} = 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 a^{2} + b^{2} = c^{2}.

∴ 2^{2} + b^{2} = 3^{2}

b^{2} = 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, c^{2} = a^{2} + b^{2}

⇒ 64 = 25 + b^{2}

⇒ b^{2} = 64 – 25

⇒ b^{2} = 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 a^{2} + b^{2} = c^{2}.

3^{2} + b^{2} = 5^{2}

⇒ b^{2} = 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, a^{2} + b^{2} = c^{2}.

∴ 4^{2} + b^{2} = 5^{2}

⇒ b^{2} = 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 a^{2} + b^{2} = c^{2}.

a^{2} + 12^{2} = 13^{2}

⇒ a^{2} = 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]

∴ b^{2} = 4a …………….(i)

We know that, c^{2} = a^{2} + b^{2}

⇒ (3√5)^{2} = a^{2} + 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

b^{2} = 5 × 4

⇒ b^{2} = 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 a^{2} = 5 and b^{2} = 20]

⇒ 20x^{2} – 25y^{2} = 500

⇒ 4x^{2} – 5y^{2} = 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

⇒ b^{2} = 6a

We know that a^{2} + b^{2} = c^{2}.

a^{2} + 6a = 16

⇒ a^{2} + 6a – 16 = 0

⇒ a^{2} + 8a – 2a – 16 = 0

⇒ (a + 8) (a – 2) = 0

⇒ a = – 8, 2

Since a is non-negative, a = 2.

b^{2} = 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 – a^{2}) – 4a^{2} = a^{2}(10 – a^{2})

⇒ 90 – 13a^{2} = 10a^{2} – a^{4}

⇒ a^{4} – 23a^{2} + 90 = a

⇒ (a^{2} – 18) (a^{2} – 5) = 0

∴ a^{2} = 18 or 5

Since, e = \(\frac{\sqrt{10}}{a}\) and e > 1

∴ \(\frac{\sqrt{10}}{a}\) > 1

⇒ a < √10; a^{2} < 10

we reject a^{2} = 18

Thus, a^{2} = 5

∴ b^{2} = a^{2} e^{2} – a^{2}

= 10 – 5 = 5

Hence, equation of hyperbola is \(\frac{y^{2}}{5}-\frac{x^{2}}{5}\) = 1 [put a^{2} = b^{2} = 5]

or y^{2} – x^{2} = 5.