NCERT Solutions for Class 11 Maths

Class 11 NCERT Solutions- Chapter 11 Conic Section – Exercise 11.3

Difficulty Level : Hard
Last Updated : 09 Mar, 2021

In each of the Exercises 1 to 9, find the coordinates of the foci, the vertices, the length of the major axis, the minor axis, the eccentricity, and the length of the latus rectum of the ellipse.

Question 1. $\frac{x^2}{36} + \frac{y^2}{16}$ = 1

Solution:

Since denominator of x²/36 is larger than the denominator of y²/16,
the major axis is along the x-axis.
On comparing the given equation with $\frac{x^2}{a^2} + \frac{y^2}{b^2}$ = 1, we get
a²= 36 and b²= 16
⇒ a = ±6 and b = ±4
The Foci:
Foci = (c, 0) and (-c, 0) when (a²> b²)
c = √(a²– b²) -(when a²> b²)
c = √(36 – 16)
c = √20 = 2√5
⇒ (2√5, 0) and (-2√5, 0)
Vertices:
Vertices = (a, 0) and (-a, 0) when (a²> b²)
⇒ (6, 0) and (-6, 0)
Length of major axis:
Length of major axis = 2a (when a²> b²)
= 2 × 6
⇒ Length of major axis = 12
Length of minor axis:
Length of minor axis = 2b (when a²> b²)
= 2 × 4
⇒ Length of minor axis = 8
Eccentricity
Eccentricity = c/a (when a²> b²)
= 2√5/6
= √5/3
Length of the latus rectum:
Length of the latus rectum = 2b²/a (when a²> b²)
= 2×16/6
= 16/3

Question 2. $\frac{x^2}{4} + \frac{y^2}{25}$ = 1

Solution:

Since denominator of y²/25 is larger than the denominator of x²/4,
the major axis is along the y-axis.
Comparing the given equation with $\frac{x^2}{a^2} + \frac{y^2}{b^2}$ = 1, we get
a²= 4 and b²= 25
⇒ a = ±2 and b = ±5
The Foci:
Foci = (0, c) and (0, -c) when (a²< b²)
c = √(b²– a²) -(when a²< b²)
c = √(25 – 4)
c = √21
⇒ (0, √21) and (0, -√21)
Vertices:
Vertices = (0, b) and (0, -b) when (a²< b²)
⇒ (0, 5) and (0, -5)
Length of major axis:
Length of major axis = 2b (when a²< b²)
= 2 × 5
⇒ Length of major axis = 10
Length of minor axis:
Length of minor axis = 2a (when a²< b²)
=2 × 2
⇒ Length of minor axis = 4
Eccentricity:
Eccentricity = c/b (when a²< b²)
= √21/5
Length of the latus rectum:
Length of the latus rectum = 2a²/b (when a²< b²)
= 2×4/5
= 8/5

Question 3. $\frac{x^2}{16} + \frac{y^2}{9}$ = 1

Solution:

Since denominator of x²/16 is larger than the denominator of y²/9,
the major axis is along the x-axis.
Comparing the given equation with $\frac{x^2}{a^2} + \frac{y^2}{b^2}$ = 1, we get
a²= 16 and b²= 9
⇒ a = ±4 and b = ±3
The Foci:
Foci = (c, 0) and (-c, 0) when (a²> b²)
c = √(a²– b²) -(when a²> b²)
c = √(16 – 9)
c = √7
⇒ (√7, 0) and (-√7, 0).
Vertices:
Vertices = (a, 0) and (-a, 0) when (a²> b²)
⇒ (4,0) and (-4,0)
Length of major axis:
Length of major axis = 2a (when a²> b²)
= 2 × 4
⇒ Length of major axis = 8
Length of minor axis:
Length of minor axis = 2b (when a²> b²)
= 2 × 3
⇒ Length of minor axis = 6
Eccentricity:
Eccentricity = c/a (when a²>b²)
= √7/4
Length of the latus rectum:
Length of the latus rectum = 2b²/a (when a²> b²)
= 2 × 9/4
= 9/2

Question 4. $\frac{x^2}{25} + \frac{y^2}{100}$ = 1

Solution:

Since denominator of y²/100 is larger than the denominator of x²/25,
the major axis is along the y-axis.
Comparing the given equation with $\frac{x^2}{a^2} + \frac{y^2}{b^2}$ = 1, we get
a²= 25 and b²= 100
⇒ a = ±5 and b = ±10
The Foci:
Foci = (0, c) and (0, -c) when (a²< b²)
c = √(b²– a²) -(when a²< b²)
c = √(100 – 25)
c = √75
c = 5√3
⇒ (0, 5√3) and (0, -5√3)
Vertices:
Vertices = (0, b) and (0, -b) when (a²< b²)
⇒ (0, 10) and (0, -10)
Length of major axis:
Length of major axis = 2b (when a²< b²)
= 2 × 10
⇒ Length of major axis = 20
Length of minor axis:
Length of minor axis = 2a (when a²< b²)
= 2 × 5
⇒ Length of minor axis = 10
Eccentricity:
Eccentricity = c/b (when a²< b²)
= 5√3/10
= √3/2
Length of the latus rectum:
Length of the latus rectum = 2a²/b (when a²< b²)
= 2 × 25/10
= 5

Question 5. $\frac{x^2}{49} + \frac{y^2}{36}$ = 1

Solution:

Since denominator of x²/49 is larger than the denominator of y²/36,
the major axis is along the x-axis.
Comparing the given equation with $\frac{x^2}{a^2} + \frac{y^2}{b^2}$ = 1, we get
a²= 49 and b²= 36
⇒ a = ±7 and b = ±6
The Foci:
Foci = (c, 0) and (-c, 0) when (a²> b²)
c = √(a²– b²) -(when a²> b²)
c = √(49 – 36)
c = √13
⇒ (√13, 0) and (-√13, 0).
Vertices:
Vertices = (a, 0) and (-a, 0) when (a²> b²)
⇒ (7, 0) and (-7, 0)
Length of major axis:
Length of major axis = 2a (when a²> b²)
= 2 × 7
⇒ Length of major axis = 14
Length of minor axis:
Length of minor axis = 2b (when a²> b²)
= 2 × 6
⇒ Length of minor axis = 12
Eccentricity:
Eccentricity = c/a (when a²> b²)
= √13/7
Length of the latus rectum:
Length of the latus rectum = 2b²/a (when a²> b²)
= 2 × 36/7
= 72/7

Question 6. $\frac{x^2}{100} + \frac{y^2}{400}$ = 1

Solution:

Since denominator of y²/400 is larger than the denominator of x²/100,
the major axis is along the y-axis.
Comparing the given equation with $\frac{x^2}{a^2} + \frac{y^2}{b^2}$ = 1, we get
a²= 100 and b²= 400
⇒ a = ±10 and b = ±20
The Foci:
Foci = (0, c) and (0, -c) when (a²< b²)
c = √(b²– a²) -(when a²< b²)
c = √(400 – 100)
c = √300
c = 10√3
⇒ (0, 10√3) and (0, -10√3)
Vertices:
Vertices = (0, b) and (0, -b) when (a²< b²)
⇒ (0, 20) and (0, -20)
Length of major axis:
Length of major axis = 2b (when a²< b²)
= 2 × 20
⇒ Length of major axis = 40
Length of minor axis:
Length of minor axis = 2a (when a²< b²)
= 2 × 10
⇒ Length of minor axis = 20
Eccentricity:
Eccentricity = c/b (when a²< b²)
= 10√3/20
= √3/2
Length of the latus rectum:
Length of the latus rectum = 2a²/b (when a²< b²)
= 2×100/20
= 10

Question 7. 36x² + 4y² = 144

Solution:

36x² + 4y² = 144
Dividing LHS and RHS by144,
$\frac{36x^2}{144} + \frac{4y^2}{144} = \frac{144}{144}$
$\frac{x^2}{4} + \frac{y^2}{36}$ = 1 (Obtained Equation)
Since denominator of y²/36 is larger than the denominator of x²/4,
the major axis is along the y-axis.
Comparing the given equation with $\frac{x^2}{a^2} + \frac{y^2}{b^2}$ = 1, we get
a²= 4 and b²= 36
⇒ a = ±2 and b = ±6
The Foci:
Foci = (0, c) and (0, -c) when (a²< b²)
c = √(b²– a²) -(when a²< b²)
c = √(36 – 4)
c = √32
c = 4√2
⇒ (0, 4√2) and (0, -4√2)
Vertices:
Vertices = (0, b) and (0, -b) when (a²< b²)
⇒ (0, 6) and (0, -6)
Length of major axis:
Length of major axis = 2b (when a²<b²)
= 2 × 6
⇒ Length of major axis = 12
Length of minor axis
Length of minor axis = 2a (when a²< b²)
= 2 × 2
⇒ Length of minor axis = 4
Eccentricity:
Eccentricity = c/b (when a²< b²)
= 4√2/6
= 2√2/3
Length of the latus rectum:
Length of the latus rectum = 2a²/b (when a²< b²)
= 2 × 4/6
= 4/3

Question 8. 16x² + y² = 16

Solution:

16x² + y² = 16
Dividing LHS and RHS by16,
$\frac{16x^2}{16} + \frac{y^2}{16} = \frac{16}{16}$
$\frac{x^2}{1} + \frac{y^2}{16}$ = 1 (Obtained Equation)
Since denominator of y²/16 is larger than the denominator of x²/1,
the major axis is along the y-axis.
Comparing the given equation with $\frac{x^2}{a^2} + \frac{y^2}{b^2}$ = 1, we get
a²= 1 and b²= 16
⇒ a = ±1 and b = ±4
The Foci:
Foci = (0, c) and (0, -c) when (a²< b²)
c = √(b²– a²) -(when a²< b²)
c = √(16 – 1)
c = √15
⇒ (0, √15) and (0, -√15)
Vertices:
Vertices = (0, b) and (0, -b) when (a²< b²)
⇒ (0, 4) and (0, -4)
Length of major axis:
Length of major axis = 2b (when a²< b²)
= 2 × 4
⇒ Length of major axis = 8
Length of minor axis:
Length of minor axis = 2a (when a²< b²)
=2 × 1
⇒ Length of minor axis = 2
Eccentricity:
Eccentricity = c/b (when a²< b²)
= √15/4
Length of the latus rectum:
Length of the latus rectum = 2a²/b (when a²< b²)
= 2 × 1/4
= 1/2

Question 9. 4x² + 9y² = 36

Solution:

4x² + 9y² = 36
Dividing LHS and RHS by 36,
$\frac{4x^2}{36} + \frac{9y^2}{36} = \frac{36}{36}$
$\frac{x^2}{9} + \frac{y^2}{4}$ = 1 (Obtained Equation)
Since denominator of x²/9 is larger than the denominator of y²/4,
the major axis is along the x-axis.
Comparing the given equation with $\frac{x^2}{a^2} + \frac{y^2}{b^2}$ = 1, we get
a² = 9 and b² = 4
⇒ a = ±3 and b = ±2
The Foci:
Foci = (c, 0) and (-c, 0) when (a² > b²)
c = √(a² – b²) -(when a² > b²)
c = √(9 – 4)
c = √5
⇒ (√5, 0) and (-√5, 0).
Vertices
Vertices = (a, 0) and (-a, 0) when (a² > b²)
⇒ (3, 0) and (-3, 0).
Length of major axis
Length of major axis = 2a (when a² > b²)
= 2 × 3
⇒ Length of major axis = 6
Length of minor axis
Length of minor axis = 2b (when a² > b²)
= 2 × 2
⇒ Length of minor axis = 4
Eccentricity
Eccentricity = c/a (when a² > b²)
= √5/3
Length of the latus rectum
Length of the latus rectum = 2b²/a (when a² > b²)
= 2 × 4/3
= 8/3

In each of the following Exercises 10 to 20, find the equation for the ellipse that satisfies the given conditions:

Question 10. Vertices (± 5, 0), foci (± 4, 0).

Solution:

Since the vertices are on x-axis, the equation will be of the form
$\frac{x^2}{a^2} + \frac{y^2}{b^2}$ = 1, where a is the semi-major axis. (where a²> b²)
Given that a = ±5, c = ±4
As, from the relation
c² = a²– b² (when a²> b²)
b² = a²– c²
b² = 25 – 16
b² = 9
So, a² = 25 and b² = 9
Hence, the required equation of ellipse,
$\frac{x^2}{25} + \frac{y^2}{9}$ = 1

Question 11. Vertices (0, ± 13), foci (0, ± 5).

Solution:

Since the vertices are on y-axis, the equation will be of the form
$\frac{x^2}{a^2} + \frac{y^2}{b^2}$ = 1, where b is the semi-major axis. (where a²< b²)
Given that b = ±13, c = ±5
As, from the relation
c² = b²– a² (when a²< b²)
a² = b²– c²
a² = 169 – 25
a² = 144
So, a² = 144 and b² = 169
Hence, the required equation of ellipse,
$\frac{x^2}{144} + \frac{y^2}{169}$ = 1

Question 12. Vertices (± 6, 0), foci (± 4, 0).

Solution:

Since the vertices are on x-axis, the equation will be of the form
$\frac{x^2}{a^2} + \frac{y^2}{b^2}$ = 1, where a is the semi-major axis. (where a²> b²)
Given that a = ±6, c = ±4
As, from the relation
c² = a²– b² (when a²> b²)
b² = a²– c²
b² = 36 – 16
b² = 20
So, a² = 36 and b² = 20
Hence, the required equation of ellipse,
$\frac{x^2}{36} + \frac{y^2}{20}$ = 1

Question 13. Ends of major axis (± 3, 0), ends of minor axis (0, ± 2).

Solution:

Since the major axis are on x-axis, and minor axis on the y-axis, the equation will be of the form
$\frac{x^2}{a^2} + \frac{y^2}{b^2}$ = 1, where a is the semi-major axis. (where a²> b²)
Given that a = ±3, b = ±2
So, a² = 9 and b² = 4
Hence, the required equation of ellipse,
$\frac{x^2}{9} + \frac{y^2}{4}$ = 1

Question 14. Ends of major axis (0, ±√5), ends of minor axis (± 1, 0).

Solution:

Since the major axis are on y-axis, and minor axis on the x-axis, the equation will be of the form
$\frac{x^2}{a^2} + \frac{y^2}{b^2}$ = 1, where b is the semi-major axis. (where a²< b²)
Given that a = ±1, b = ±√5
So, a² = 1 and b² = 5
Hence, the required equation of ellipse,
$\frac{x^2}{1} + \frac{y^2}{5}$ = 1

Question 15. Length of major axis 26, foci (± 5, 0).

Solution:

Since the foci are on x-axis, the equation will be of the form
$\frac{x^2}{a^2} + \frac{y^2}{b^2}$ = 1, where a is the semi-major axis. (where a²> b²)
Given that c = ±5 and Length of major axis = 26
As, Length of major axis = 2a (when a² > b²)
2a = 26
a = 13
As, from the relation
c² = a²– b² (when a²> b²)
b² = a²– c²
b² = 169 – 25
b² = 144
So, a² = 169 and b² = 144
Hence, the required equation of ellipse,
$\frac{x^2}{169} + \frac{y^2}{144}$ = 1

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

Solution:

Since the foci are on y-axis, the equation will be of the form
$\frac{x^2}{a^2} + \frac{y^2}{b^2}$ = 1, where b is the semi-major axis. (where a²< b²)
Given that c = ±6 and Length of minor axis = 16
As, Length of minor axis = 2a (when a²< b²)
2a = 16
a = 8
As, from the relation
c² = b²– a² (when a²< b²)
b² = c²+ a²
b² = 36 + 64
b² = 100
So, a² = 64 and b² = 100
Hence, the required equation of ellipse,
$\frac{x^2}{64} + \frac{y^2}{100}$ = 1

Question 17. Foci (± 3, 0), a = 4.

Solution:

Since the foci are on x-axis, the equation will be of the form
$\frac{x^2}{a^2} + \frac{y^2}{b^2}$ = 1, where a is the semi-major axis. (where a²> b²)
Given that a = 4 and c = ±3
As, from the relation
c² = a²– b² (when a²> b²)
b² = a²– c²
b² = 16 – 9
b² = 7
So, a² = 16 and b² = 7
Hence, the required equation of ellipse,
$\frac{x^2}{16} + \frac{y^2}{7}$ = 1

Question 18. b = 3, c = 4, centre at the origin; foci on the x axis.

Solution:

Since the foci are on x-axis, the equation will be of the form
$\frac{x^2}{a^2} + \frac{y^2}{b^2}$ = 1, where a is the semi-major axis. (where a²> b²)
Given that b = 3 and c = 4
As, from the relation
c² = a²– b² (when a²> b²)
a² = b²+ c²
a² = 9 + 16
a² = 25
So, a² = 25 and b² = 9
Hence, the required equation of ellipse,
$\frac{x^2}{25} + \frac{y^2}{9}$ = 1

Question 19. Centre at (0,0), major axis on the y-axis and passes through the points (3, 2) and (1, 6).

Solution:

The standard equation of ellipse having centre (0, 0) will be of the form
$\frac{x^2}{a^2} + \frac{y^2}{b^2}$ = 1
Since the points (3, 2) and (1, 6) lie on the ellipse, we can have
$\frac{3^2}{a^2} + \frac{2^2}{b^2}$ = 1
$\frac{9}{a^2} + \frac{4}{b^2}$ = 1 -(1)
$\frac{1^2}{a^2} + \frac{6^2}{b^2}$ = 1
$\frac{1}{a^2} + \frac{36}{b^2}$ = 1 -(2)
Eq(2) subtracted from (multiplying eq(1) by 9) and, we get
9×( $\frac{9}{a^2} + \frac{4}{b^2}$ ) – ( $\frac{1}{a^2} + \frac{36}{b^2}$ ) = 9 – 1
$\frac{81}{a^2} + \frac{36}{b^2} - \frac{1}{a^2} - \frac{36}{b^2}$ = 8
80/a² = 8
a² = 80/8
a² = 10
Now, substituting a² = 10 in eq(1)
$\frac{9}{a^2} + \frac{4}{b^2}$ = 1
9/10 + 4/b² = 1
4/b² = 1 – 9/10
4/b² = 1/10
b² = 10 × 4 = 40
So, a² = 10 and b² = 40
Hence, the required equation of ellipse,
$\frac{x^2}{10} + \frac{y^2}{40}$ = 1

Question 20. Major axis on the x-axis and passes through the points (4, 3) and (6, 2).

Solution:

The standard equation of ellipse having centre (0, 0) will be of the form
$\frac{x^2}{a^2} + \frac{y^2}{b^2}$ = 1
Since the points (4,3) and (6,2) lie on the ellipse, we can have
$\frac{4^2}{a^2} + \frac{3^2}{b^2}$ = 1
$\frac{16}{a^2} + \frac{9}{b^2}$ = 1 -(1)
and, $\frac{6^2}{a^2} + \frac{2^2}{b^2}$ = 1
$\frac{36}{a^2} + \frac{4}{b^2}$ = 1 -(2)
(multiplying eq(2) by 9) subtracted from (multiplying eq(1) by 4) and, we get
9×( $\frac{36}{a^2} + \frac{4}{b^2}$ ) – 4×( $\frac{16}{a^2} + \frac{9}{b^2}$ ) = 9 – 4
$\frac{324}{a^2} + \frac{36}{b^2} - \frac{64}{a^2} - \frac{36}{b^2}$ = 5
260/a² = 5
a² = 260/5
a² = 52
Now, substituting a² = 52 in eq(1)
$\frac{16}{2} + \frac{9}{b^2}$ = 1
9/b² = 1 – 16/52
9/b² = 36/52
9/b² = 36/52
b² = 9 × 36/52 = 13
So, a² = 52 and b² = 13
Hence, the required equation of ellipse,
$\frac{x^2}{52} + \frac{y^2}{13}$ = 1