# Class 11 Maths NCERT Solutions for Chapter 11 Conic Sections Exercise 11.3

### Conic Sections Exercise 11.3 Solutions

1. 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 x2/36 + y2/16 = 1

Solution

The given equation is x2/36 + y2/16 = 1.
Here, the denominator of x2/36 is greater than the denominator of y2/16 .
Therefore, the major axis is along the x - axis, while the minor axis is along the y - axis.
On comparing the given equation with x2/a2 + y2/b2  = 1 , we obtain a = 6 and b = 4.

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

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 x2/4 + y2/25 = 1

Solution

The given equation is
Here, the denominator of y2 /25 is greater than the denominator of x2 /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  , we obtain b = 2 and a  = 5.

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

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 x2/16 + y2/9 = 1.

Solution

The given equation is
Here, the denominator of x2 /16 is greater than the denominator of y2 /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 x2/a2 + y2/b2  = 1, we obtain a = 4 and b = 3.

Therefore,
The coordinates of the foci are (±√7 , 0).
The coordinates of the vertices are (±4, 0).
Length of major axis = 2a = 8
Length of minor axis = 2b = 6

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 x2/25 + y2/100 = 1

Solution

The given equation is
Here, the denominator of y2/100 is greater than the denominator of x2/25.
Therefore, the major axis is along the y - axis, while the minor axis is along the x-axis.
On comparing the given equation with x2 /b2  + y2 /a2  = 1, we obtain b = 5 and a = 10.

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

5. 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 x2/49 + y2/36 = 1

Solution

The given equation is
Here, the denominator of x2/49 is greater than the denominator of y2/36.
Therefore, the major axis is along the x - axis, while the minor axis is along the y - axis.
On comparing the given equation with x2/a2 + y2/b2  = 1, we obtain a = 7 and b = 6.

Therefore,
The coordinates of the foci are (± √13, 0) .
The coordinates of the vertices are (±7, 0).
Length of major axis = 2a = 14
Length of minor axis = 2b = 12

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 x2/100 + y2/400 = 1

Solution

The given equation is
Here, the denominator of y2 /400 is greater than the denominator of x2/100.
Therefore, the major axis is along the y-axis, while the minor axis is along the x-axis.
On comparing the given equation with x2/b2 + y2/a2 = 1, we obtain b = 10 and a = 20.

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

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 36x2 + 4y2 = 144

Solution

The given equation is 36x2 + 4y2 = 144.
It can be written as

Here, the denominator of y2/62 is greater than the denominator of x2/22.
Therefore, the major axis is along the y - axis, while the minor axis is along the x - axis.
On comparing equation (1) with x2/b2 + y2/a2 = 1, we obtain b = 2 and a = 6.

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

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 ellipse 16x2 + y2 = 16

Solution

The given equation is 16x2 + y2 = 16.
It can be written as

Here, the denominator of y2/42 is greater than the denominator of x2/12.
Therefore, the major axis is along the y - axis, while the minor axis is along the x - axis.
On comparing equation (1) with x2/b2 + y2/a2 = 1, we obtain b = 1 and a = 4.

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
Eccentricity, e = c/a = √15/4
Length of latus rectum - 2b2/a = (2×1)/4 = 1/2

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 4x2 + 9y2 = 36

Solution

The given equation is 4x2 + 9y2 = 36.
It can be written as

Here, the denominator of x2/32 is greater than the denominator of y2/22 .
Therefore, the major axis is along the x - axis, while the minor axis is along the y - axis.
On comparing the given equation with x2/a2 + y2/b2 = , we obtain a = 3 and b = 2.

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 = c/a = √5/3
Length of latus rectum = 2b2/a = (2× 4)/3 = 8/3

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

Solution

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 x2/a2 + y2/b2 = 1 , where is the semi-major axis.
Accordingly, a = 5 and c = 4.
It is known that a2 = b2 + c2.
∴ 52 = b2 + 42
⇒ 25 = b2 + 16
⇒ b2 = 25 - 16
⇒ b = √9 = 3
Thus, the equation of the ellipse

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

Solution

Vertices (0, ±13), foci (0, ±5)
Here, the vertices are on the y-axis.
Therefore, the equation of the ellipse will be of the form x2 /a2 + y2 /b2 = 1, where is the semi-major axis.
Accordingly, a = 13 and c = 5.
It is known that  a2 = b2 + c2.
∴ 132 = b2 + 52
⇒ 169 = b2 + 25
⇒ b2 = 169 - 25
⇒ b = √144 = 12
Thus, the equation of the ellipse is

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

Solution

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 x2 /a2 + y2 /b2 = 1, where is the semi-major axis.
Accordingly, a = 6, c = 4.
It is known that a2 = b2 + c2.
∴ 62 = b2 + 42
⇒ 36 = b2 + 16
⇒ b2 = 36 - 16
⇒ b = √ 20
Thus, the equation of the ellipse is

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

Solution

Ends of major axis (±3, 0), ends of minor axis (0, ±2)
Here, the major axis is along the x-axis.
Therefore, the equation of the ellipse will be of the form x2/a2 + y2/b2 = 1, where a is the semi-major axis.
Accordingly, a = 3 and b = 2.
Thus, the equation of the ellipse is

14. Find the equation for the ellipse that satisfies the given conditions: Ends of major axis (0, ± √5), ends of minor axis (±1, 0)

Solution

Ends of major axis (0, ± √5), ends of minor axis (±1, 0)
Here, the major axis is along the y-axis.
Therefore, the equation of the ellipse will be of the form x2 /a2 + y2 /b2 = 1, where is the semi-major axis.
Accordingly, a = √5 and b = 1.
Thus, the equation of the ellipse is

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

Solution

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  x2 /a2 + y2 /b2 = 1, where is the semi-major axis.
Accordingly,
2a = 26
⇒ a = 13 and c = 5.
It is known that a2 = b2 + c2.
∴ 132 = b2 + 52
⇒ 169 = b2 + 25
⇒ b2 = 169 - 25
⇒ b = √144 = 12
Thus, the equation of the ellipse is

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

Solution

Length of minor axis = 16; foci = (0, ± 6).
Since the foci are on the y-axis, the major axis is along the y-axis.
Therefore, the equation of the ellipse will be of the form x2 /a2 + y2 /b2 = 1, where a is the semi-major axis.
Accordingly,
2b = 16
⇒ b = 8 and c = 6.
It is known that a2 = b2 + c2.
∴ a2 = 82 + 62 = 64 + 36 = 100
⇒ a = √100 = 10
Thus, the equation of the ellipse is

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

Solution

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 x2/a2 + y2/b2 = 1, where a is the semi-major axis.
Accordingly, c = 3 and a = 4.
It is known that a2 = b2 + c2.
∴ 42 = b2 + 32
⇒ 16 = b2 + 9
⇒ b2 = 16 - 9 = 7
Thus, the equation of the ellipse is x2/16 + y2/7 = 1.

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

Solution

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 x2/a2 + y2/b2 = 1, where a is the semi-major axis.
Accordingly, b = 3, c = 4.
It is known that a2 = b2 + c2.
∴ a2 = 32 + 42 = 9 + 16 = 25
⇒ a = 5
Thus, the equation of the ellipse is

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).

Solution

Since the centre is at (0, 0) and the major axis is on the y-axis, the equation of the ellipse will be of the form

Where, a is the semi - major axis
The ellipse passes through points (3, 2) and (1, 6), Hence,

On solving equations (2) and (3), we obtain b2 = 10 and a2 = 40.
Thus, the equation of the ellipse is x2/10 + y2/40 = 1 or 4x2 + y2 = 40.

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).

Solution

Since the major axis is on the x - axis, the equation of the ellipse will be of the form

Where, a is the semi - major axis
The ellipse passes through points (4, 3) and (6, 2). Hence,

On solving equations (2) and (3), we obtain a2 = 52 and b2 = 13.
Thus, the equation of the ellipse is x2/52 + y2/13 = 1 or x2 + 4y2 = 52.