Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 11 Conic Sections Ex 11.3 Textbook Exercise Questions and Answers.

## PSEB Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.3

Question 1.

Find the coordinates of the foci, the vertices, the length of mqjor axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse \(\frac{x^{2}}{36}+\frac{y^{2}}{16}\) = 1.

Answer.

The given equation is \(\frac{x^{2}}{36}+\frac{y^{2}}{16}\) = 1.

Here, the denominator of \(\frac{x^{2}}{36}\) is greater than the denominator of \(\frac{y^{2}}{16}\).

Therefore, the major axis is along the x-axis, while the minor axis is along the y-axis.

On comparing the given equation with \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1, we otain a = 6 and b = 4.

c = \(\sqrt{a^{2}-b^{2}}=\sqrt{36-16}\)

= √20 = 2√5.

Therefore,

The coordinates of the foci are (2√5, 0) and (- 2√5, 0).

The coordinates of the vertices are (6, 0) and (- 6, 0).

Length of major axis = 2a = 12.

Length of minor axis = 2b = 8

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

Length of latus rectum = \(\frac{2 b^{2}}{a}=\frac{2 \times 16}{6}=\frac{16}{3}\).

Question 2.

Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse \(\frac{x^{2}}{4}+\frac{y^{2}}{25}\) = 1.

Answer.

The given equation is \(\frac{x^{2}}{4}+\frac{y^{2}}{25}\) = 1 or \(\frac{x^{2}}{2^{2}}+\frac{y^{2}}{5^{5}}\) = 1.

Here, the denominator of \(\frac{y^{2}}{25}\) is greater than the denominator of \(\frac{x^{2}}{4}\).

Therefore, the major axis is along the y-axis, while the minor axis is along the x-axis.

On comparing the given equation with \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}\)= 1, we obtain

b = 2 and a = 5

∴ c = \(\sqrt{a^{2}-b^{2}}=\sqrt{25-4}=\sqrt{21}\)

Therefore, the coordinates of the foci are (0, √21) and (0, – √21).

The coordinates of the vertices are (0, 5) and (0, – 5).

Length of major axis = 2a = 10

Length of minor axis = 2b = 4

Eccentricity e = \(\frac{c}{a}=\frac{\sqrt{21}}{5}\)

Length of latus rectum = \(\frac{2 b^{2}}{a}=\frac{2 \times 4}{5}=\frac{8}{5}\)

Question 3.

Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse \(\frac{x^{2}}{16}+\frac{y^{2}}{9}\) = 1.

Answer.

The given equation is \(\frac{x^{2}}{16}+\frac{y^{2}}{9}\) = 1 or \(\frac{x^{2}}{4^{2}}+\frac{y^{2}}{3^{2}}\) = 1

Here, the denominator of \(\frac{x^{2}}{16}\) is greater than the denominator of \(\frac{y^{2}}{9}\).

Therefore, the major axis is along the x-axis, while the minor axis is along the y-axis.

On comparing the given equation with \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1 , we obtain a = 4 and b = 3.

c = \(\sqrt{a^{2}-b^{2}}=\sqrt{16-9}=\sqrt{7}\)

Therefore, the coordinates of the foci are (± c, 0) i.e., (± √7, 0).

The coordinates of the vertices are (± a, 0) i.e., (± 4, 0).

Length of major axis = 2a = 8

Length of minor axis = 2b = 6

Eccentricity e = \(\frac{c}{a}=\frac{\sqrt{7}}{4}\)

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

Question 4.

Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse \(\frac{x^{2}}{25}+\frac{y^{2}}{100}\) = 1.

Answer.

The given equation is \(\frac{x^{2}}{25}+\frac{y^{2}}{100}\) = 1 or \(\) = 1

Here, the denominator of \(\frac{y^{2}}{100}\) is greater than the denominator of \(\frac{x^{2}}{25}\)

Therefore, the major axis is along the y-axis, while the minor axis is along the x-axis.

On comparing the given equation with \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}\) = 1 , we obtain

b = 5 and a = 10.

∴ c = \(\sqrt{a^{2}-b^{2}}=\sqrt{100-25}=\sqrt{75}=5 \sqrt{3} .\)

Therefore, the coordinates of the foci are (0, ± 5√3).

The coordinates of the vertices are (0, ± 10).

Length of major axis = 2a = 20

Length of minor axis = 2b = 10

Eccentncity, e = \(\frac{c}{a}=\frac{5 \sqrt{3}}{10}=\frac{\sqrt{3}}{2}\)

Length of lattis rectum = \(\frac{2 b^{2}}{a}=\frac{2 \times 25}{10}\)

Question 5.

Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the lattis rectum of the ellipse \(\frac{x^{2}}{49}+\frac{y^{2}}{36}\) = 1.

Answer.

The given equation is \(\frac{x^{2}}{49}+\frac{y^{2}}{36}\) = 1 or \(\frac{x^{2}}{7^{2}}+\frac{y^{2}}{6^{2}}\) = 1.

Here, the denominator of \(\frac{x^{2}}{49}\) is greater than the denominator of \(\frac{y^{2}}{36}\).

Therefore, the major axis is along the x-axis, while the minor axis is along the y-axis.

On comparing the given equation with \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1 , we obtain a = 4 and b = 3.

∴ c = \(\sqrt{a^{2}-b^{2}}=\sqrt{49-36}=\sqrt{13}\)

Therefore, the coordinates of the foci are (± c, 0)

The coordinates of the vertices are (± a, 0) i.e., (± √13, 0)

Length of major axis = 2a = 14

Length of minor axis = 2b = 12

Eccentricity, e = \(\frac{c}{a}=\frac{\sqrt{13}}{7}\)

Length of latus rectum = \(\frac{2 b^{2}}{a}=\frac{2 \times 36}{7}=\frac{72}{7}\).

Question 6.

Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse \(\frac{x^{2}}{100}+\frac{y^{2}}{400}\) = 1.

Answer.

The given equation is \(\frac{x^{2}}{100}+\frac{y^{2}}{400}\) = 1 or + \(\frac{x^{2}}{10^{2}}+\frac{y^{2}}{20^{2}}\) = 1

Here, the denominator of \(\frac{y^{2}}{400}\) is greater than the denominator of \(\frac{x^{2}}{100}\).

Therefore, the major axis is along the y-axis, while the minor axis is along the x-axis.

On comparing the given equation with \(\sqrt{a^{2}-b^{2}}=\sqrt{400-100}=\sqrt{300}=10 \sqrt{3}\) = 1 , we obtain b = 10 and a = 20

∴ c = \(\sqrt{a^{2}-b^{2}}=\sqrt{400-100}=\sqrt{300}=10 \sqrt{3}\)

Therefore, the coordinates of the foci are (0, ± 10√3)

The coordinates of the vertices are (0, ± 20)

Length of major axis = 2a = 40

Length of minor axis = 2b = 20

Eccentricity, e = \(\frac{c}{a}=\frac{10 \sqrt{3}}{20}=\frac{\sqrt{3}}{2}\)

Length of latus recrum = \(\frac{2 b^{2}}{a}=\frac{2 \times 100}{20}\) =10.

Question 7.

Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse 36x^{2} + 4y^{2} = 144.

Answer.

The given equation is 36x^{2} + 4y^{2} = 144

It can be writen as 36x^{2} + 4y^{2} = 144

or \(\frac{x^{2}}{4}+\frac{y^{2}}{36}\) = 1

or \(\frac{x^{2}}{2^{2}}+\frac{y^{2}}{6^{2}}\) = 1

Here, the denominator of \(\frac{y^{2}}{6^{2}}\) is greater than the denominator of \(\frac{x^{2}}{2^{2}}\)

Therefore, the major axis is along the y-axis, while the minor axis is along the x-axis.

On comparing equation (i) with \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}\) = 1, we obtain b = 2 and a = 6.

∴ c = \(\sqrt{a^{2}-b^{2}}=\sqrt{36-4}=\sqrt{32}=4 \sqrt{2}\)

Therefore, The coordinates of the foci are (0, ± 4√2)

The coordinates of the vertices are (0, ± 6)

Length of major axis = 2a = 12

Length of minor axis = 2b = 4

Eccentricity, e = \(\frac{c}{a}=\frac{4 \sqrt{2}}{6}=\frac{2 \sqrt{2}}{3}\)

Length of latus recrum = \(\frac{2 b^{2}}{a}=\frac{2 \times 4}{6}=\frac{4}{3}\).

Question 8.

Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the effipse 16x^{2} + y^{2} = 16.

Answer.

The given equation is 16x^{2} + y^{2} = 16

It can be writen as 16x^{2} + y^{2} = 16

or \(\frac{x^{2}}{1}+\frac{y^{2}}{16}\)

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

Here, the denominator of \(\frac{y^{2}}{4^{2}}\) is greater than the denominator of \(\frac{x^{2}}{1^{2}}\).

Therefore, the major axis is along the y-axis, while the minor axis is along the x-axis.

On comparing the equation (i) with \(\) = 1, we obtain b = 1 and a = 4.

∴ c = \(\sqrt{a^{2}-b^{2}}=\sqrt{16-1}=\sqrt{15}\)

Therefore, the coordinates of the foci are (0, ± √15)

The coordinates of the vertices are (0, ± 4)

Length of major axis = 2a = 8

Length of minor axis = 2b = 2 c V15

Eccentncity e = \(\frac{c}{a}=\frac{\sqrt{15}}{4}\)

Length of latus rectum = \(\frac{2 b^{2}}{a}=\frac{2 \times 1}{4}=\frac{1}{2}\)

Question 9.

Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse 4x^{2} + 9y^{2} = 36.

Answer.

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

It can be writen as 4x^{2} + 9y^{2} = 36

or \(\frac{x^{2}}{9}+\frac{y^{2}}{4}\) = 1

or \(\frac{x^{2}}{3^{2}}+\frac{y^{2}}{2^{2}}\) = 1 ……………..(i)

Here, the denominator of \(\frac{x^{2}}{3^{2}}\) is greater than the denominator of \(\frac{y^{2}}{2^{2}}\).

Therefore, the major axis is along the x-axis, while the minor axis is along they-axis.

On comparing the given equation with \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1, we obtain

a = 3 and b = 2.

∴ c = \(\sqrt{a^{2}-b^{2}}=\sqrt{9-4}=\sqrt{5}\)

Therefore, the coordinates of the foci are (± √5, 0)

The coordinates of the vertices are (± 3,0)

Length of major axis = 2a = 6

Length of minor axis = 2b = 4

Eccentricity, e = \(\frac{c}{a}=\frac{\sqrt{5}}{3}\)

Length of latus rectum = \(\frac{2 b^{2}}{a}=\frac{2 \times 4}{3}=\frac{8}{3}\).

Question 10.

Find the equation for the ellipse that satisfies the given conditions: Vertices (± 5, 0), foci (± 4, 0).

Answer.

Given, vertices (± 5, 0), foci (± 4 ,0)

Here, the vertices are on the x-axis.

Therefore, 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, a = 5 and c = 4.

It is known that, a^{2} = b^{2} + c^{2}

∴ 5^{2} = b^{2} + 4^{2}

⇒ 25 = b^{2} + 16

⇒ b^{2} = 25 – 16 = 9

b = √9 = 3.

Thus, the equation of the ellipse is

\(\frac{x^{2}}{5^{2}}+\frac{y^{2}}{3^{2}}\) = 1 or \(\frac{x^{2}}{25}+\frac{y^{2}}{9}\) = 1.

Question 11.

Find the equation for the ellipse that satisfies the given conditions: Vertices (0, ± 13), foci (0, ± 5).

Answer.

Given, vertices (0, ± 13), foci (0, ± 5)

Here, the vertices are on the y-axis.

Conic Sections

Therefore, the equation of the ellipse will be of the form \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}\) = 1,

where a is the semi-major axis.

Accordingly, a = 13 and c = 5.

It is known that, a^{2} = b^{2} + c^{2}

13^{2} = b^{2} + 5^{2}

⇒169 = ^{2} + 25

⇒ b^{2} = 169 – 25

⇒ b = √144 = 12

Thus, the equation of the ellipse is

\(\frac{x^{2}}{12^{2}}+\frac{y^{2}}{13^{2}}\) = 1 or

\(\frac{x^{2}}{144}+\frac{y^{2}}{169}\) = 1.

Question 12.

Find the equation for the ellipse that satisfies the given conditions: Vertices (± 6, 0), foci (± 4, 0).

Answer.

Given, vertices (± 6, 0), foci (± 4, 0)

Here, the vertices are on the x-axis.

Therefore, 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, a = 6 and c = 4

It is known that,

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

⇒ 6^{2} = b^{2} + 4^{2}

⇒ 36 = b^{2} + 16

⇒ b^{2} = 36 – 16

⇒ b = √20

Thus, the equation of the ellipse is \(\frac{x^{2}}{6^{2}}+\frac{y^{2}}{(\sqrt{20})^{2}}\) or \(\frac{x^{2}}{36}+\frac{y^{2}}{20}\) = 1.

Question 13.

Find the equation for the ellipse that satisfies the given conditions: Ends of major axis (± 3, 0), ends of minor axis (0, ± 2).

Answer.

Given, ends of major axis (± 3, 0), ends of monor axis (0, ± 2).

Here, the major axis is along the x-axis.

Therefore, 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, a = 3 and b = 2

Thus, the equation of the ellipse is \(\frac{x^{2}}{3^{2}}+\frac{y^{2}}{2^{2}}\) i.e., \(\frac{x^{2}}{9}+\frac{y^{2}}{4}\) = 1.

Question 14.

Find the equation for the ellipse that satisfies the given conditions: Ends of major axis (0, ±V5), ends of minor axis (± 1, 0).

Answer.

Given, ends of major axis (0, ± √5), ends of minor axis (± 1, 0).

Here, the vertices are on the y-axis.

Therefore, the equation of the ellipse will be of the form \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}\) = 1,

where a is the semi-major axis.

Accordingly, a = √5 and b = 1

Thus the equation of the ellipse is \(\frac{x^{2}}{1^{2}}+\frac{y^{2}}{(\sqrt{5})^{2}}\) = 1 or \(\frac{x^{2}}{1}+\frac{y^{2}}{5}\) = 1.

Question 15.

Find the equation for the ellipse that satisfies the given conditions: Length of major axis 26, foci (+5,0).

Answer.

Given, length of major axis = 26, foci (± 5, 0)

Since the foci are on the x-axis, the major axis is along the x-axis.

Therefore, 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 = 26

⇒ a = 13 and c = 5

It is known that a^{2} = b^{2} + c^{2}

Thus, the equation of the ellipse is \(\frac{x^{2}}{13^{2}}+\frac{y^{2}}{12^{2}}\) = 1 or \(\frac{x^{2}}{169}+\frac{y^{2}}{144}\) = 1.

Question 16.

Find the equation for the ellipse that satisfies the given conditions: Length of major axis 16, foci (0, ± 6).

Answer.

Length of minor axis = 16, foci (0, ± 6)

Since the foci are on the y-axis, the major axis is along they-axis.

Therefore, the equation of the ellipse will be of the form \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}\) = 1,

where a is the semi-major axis.

Accordingly, 2b = 16

⇒ b = 8 and c = 6

It is known that a^{2} = b^{2} + c^{2}

∴ a^{2} = 8^{2} + 6^{2}

= 64 + 36 = 100

a = √100 = 10

Thus, the equation of the ellipse is \(\frac{x^{2}}{8^{2}}+\frac{y^{2}}{10^{2}}\) = 1 or \(\frac{x^{2}}{64}+\frac{y^{2}}{100}\) = 1.

Question 17.

Find the equation for the ellipse that satisfies the given conditions: foci (± 3, 0), a = 4.

Answer.

Given, foci (± 3, 0), a = 4

Since the foci are on the x-axis, the major axis is along the x-axis.

Therefore, the equation of the ellipse will be of the form \(\) = 1,

where a is the semi-major axis.

Accordingly, c = 3 and a = 4

It is known that, a^{2} = b^{2} + c^{2}

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

⇒ 16 = b^{2} + 9

⇒ b^{2} = 16 – 9 = 7

Thus, the equation of the ellipse is \(\frac{x^{2}}{16}+\frac{y^{2}}{7}\) = 1.

Question 18.

Find the equation for the ellipse that satisfies the given conditions: b = 3, c = 4, centre at the origin; foci on the x- axis. Answer.

It is given that b = 3, c = 4, centre at the origin; foci on the x-axis.

Since the foci are on the x-axis, the major axis is along the x-axis.

Therefore, the equation of the ellipse will be of the form \(\) = 1,

where a is the semi-major axis.

Accordingly, b = 3 and c = 4

It is known that a^{2} = b^{2} + c^{2}

∴ a^{2} = 3^{2} + 4^{2}

= 9 + 16 = 25

⇒ a = 5

Thus, the equation of the ellipse is \(\frac{x^{2}}{5^{2}}+\frac{y^{2}}{3^{2}}\) = 1 or \(\frac{x^{2}}{25}+\frac{y^{2}}{9}\) = 1.

Question 19.

Find the equation for the ellipse that satisfies the given conditions: Centre at (0, 0), major axis on the y-axis and passes through the points (3, 2) and (1, 6).

Answer.

Since, major axis is along Y-axis, so equation of ellipse is

\(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}\) = 1, a > b ……………(i)

Thus, the equation of ellipse is passing through the points (3, 2) and (1, 6).

So, point (3, 2) lies on eq. (i), gives \(\frac{9}{b^{2}}+\frac{4}{a^{2}}\) = 1 ……………(ii)

Also, point (1, 6) lies on eq. (i), therefore

\(\frac{1}{b^{2}}+\frac{36}{a^{2}}\) = 1

\(\frac{9}{b^{2}}+\frac{324}{a^{2}}\) = 9

On subtracting eq. (ii) from eq. (iii), we get

\(\frac{20}{a^{2}}\) = 8

a^{2} = \(\frac{320}{8}\) = 4o

On putting the value of a^{2} in eq. (ii), we get

\(\frac{9}{b^{2}}+\frac{4}{40}\) = 1

⇒ \(\frac{9}{b^{2}}=1-\frac{1}{10}=\frac{9}{10}\)

⇒ b^{2} = 10

On putting the values of a and b in eq. (i), we get \(\frac{x^{2}}{10}+\frac{y^{2}}{40}\) = 1.

Question 20.

Find the equation for the ellipse that satisfies the given conditions: Major axis on the x-axis and passes through the points (4, 3) and (6, 2).

Answer.

Let the major axis is along X-axis, then the equation of ellipse be \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1 ………………(i)

Since, the equation of ellipse is passing through the points (4, 3) and (6, 3).

So, point (4, 3) lies on it.

\(\frac{16}{a^{2}}+\frac{9}{b^{2}}\) = 1 ……………..(ii)

Also, point (6, 2) lies on it.

\(\frac{36}{a^{2}}+\frac{4}{b^{2}}\) = 1 ………………..(iii)

On multiplying eq. (ii) by 4 and eq. (iii) by 9 and subtracting eq. (ii) from eq. (iii), we get

\(\frac{324}{a^{2}}-\frac{64}{a^{2}}\) = 9 – 4

⇒ \(\frac{260}{a^{2}}\) = 5

⇒ a^{2} = \(\frac{260}{5}\) = 52

∴ a^{2} = 52

On putting the value of a^{2} in eq. (ii), we get

\(\frac{16}{52}+\frac{9}{b^{2}}\) = 1

⇒ \(\frac{9}{b^{2}}=1-\frac{16}{52}=\frac{36}{52}\)

⇒ b^{2} = 13

Hence, equation of ellipse is \(\frac{x^{2}}{52}+\frac{y^{2}}{13}\) = 1 [put a^{2} = 52 and b^{2} = 13].