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

If you want NCERT Solutions of Chapter 11 Conic Sections Exercise 11.3 then you can find them here. These Class 11 Maths NCERT Solutions will guide you in covering all the important points and concepts of the chapter easily. These NCERT Solutions are updated as per the latest pattern released by CBSE.

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.

Since denominator of x 2 is larger than the denominator of y2, the major axis is along the x-axis. Comparing the given equation with
x2/a2 + y2/b2 = 1, we get a = 6 and b = 4.
Also √a2 - b2 = √36 - 16 = √20
Therefore, the coordinates of the foci, (–c, 0) and (c, 0) are (-√20, 0) and (√20, 0). - Vertices, (–a, 0) and (a, 0) are (–6, 0) and (6, 0). Length of the major axis, 2a is 12 units.
The length of the minor axis, 2b is 8 units and eccentricity, e = c/a is √20/6 = √5/3
The length of latus rectum = 2b2/a = 2(4)2/6
= 16/3 units.

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

Since the denominator of y 2 is larger than the denominator of x 2, the major axis is along the y-axis. Comparing the given equation with the standard equation
x2/b2 + y2/a2 = 1
We have b = 2 and a = 5
Also c =√a2 - b2 = √25 - 4 = √21
and e = c/a = √21/5
Hence, the foci, (0, c) and (0, –c) are (0, √21) and (0, – √21); the vertices are (0, 5) and (0, –5); length of the major axis, 2a is 10 units, the length of the minor axis, 2b is 4 units and the eccentricity of the ellipse, e is √21/5 . The length of latus rectum = 2b2/a = 8/5 units.

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

Equation of ellipse is x2/16 + y2/9 = 1
Here, a 2 = 16, ∴ a = 4 and b 2 = 9, ∴ b = 3 Major axis is along x-axis,
c2 = a2 – b2 = 16 – 9 = 7 ∴ c = √7
Co-ordinates of foci (± c, 0) i.e., (±√7, 0)
Vertices are (± a, 0) i.e., (± 4, 0)
Length of major axis = 2a = 2 × 4 = 8
Length of minor axis = 2b = 2 × 3 = 6
e = c/a = √7/4
Latus rectum = 2b2/a = (2 × 9)/4 = 9/2

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

Equation of ellipse is x2/25 + y2/100 = 1
Major axis is along y-axis
a 2 = 100, ∴ a = 10, b 2 = 25 ∴ b = 5
c2 = a2 – b2 = 100 – 25 = 75 ∴ c = 5√3
Foci are (0, ± c) i.e., (0, ± 5√3)
Vertices are (0, ± a) i.e., (0, ± 10)
Length of major axis = 2a = 2 × 10 = 20
Length of minor axis = 2b = 2 × 5 = 10
e = c/a = 5√3/10 = √3/2
Length of Latus rectum = 2b2/a = (2 × 25)/10 = 5

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

x2/49 + y2/36 = 1 is the equation of ellipse.
a 2 = 49, major axis is along x-axis
a 2 = 49 ∴ a = 7, b 2 = 36, b = 6
∴ c2 = a2 – b2 = 49 – 36 = 13 c = √13
Foci are (± c, 0) i.e., (± √13, 0)
Vertices are (± a, 0) i.e., (± 7, 0)
Length of major axis = 2b = 2 × 7 = 14
Length of minor axis = 2b = 2 × 6 = 12
Length of Latus rectum = 2b2/a = (2 × 36)/7 = 72/7
Ecentricity, e = c/a = √13/7

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

x2/ 100 + y2/400 = 1 is the equation of ellipse.
Major axis is along y-axis
a 2 = 400, ∴ a = 20, b 2 = 100 ∴ b = 10
c2 = a2 – b2 = 400 – 100 = 300 ∴ c = 10√3
Vertices are (0, ± a) i.e., (0, ± 20)
∴ Foci are (0, ± c) i.e., (0, ±10√3)
Length of major axis = 2a = 2 × 20 = 40
Length of minor axis = 2b = 2 × 10 = 20
Eccentricity, e = c/a = 10√3/20 = √3/2
Length of Latus rectum = 2b2/a = (2 × 100)/20 = 10

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

The given equation of the ellipse can be written if standard form as
36x2/144 + 4y2/144 = 1
Or x2/4 + y2/36 = 1
Since the denominator of x2 < denominator of y 2, the major axis is along the y-axis. Comparing the given equation with the standard equation
x2/b2 + y2/a2 = 1,
we have b = 2 and a = 6.
Also, c =√a - b =√36 - 4 =√32 = 4√2
e = c/a = √34/6 = 4√2/6 = 2√2/3
Hence, the foci, are (0, 4√2) and (0, – 4√2). Vertices, are (0, 6) and (0, –6); the length of the major axis, 2a is 12 units, the length of the minor axis, 2b is 4 units and the eccentricity of the ellipse, e is 2√2/3.
The length of latus rectum,
2b2/a is (2(22))/6 = 4/3 units.

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 16x 2 + y 2 = 16.

The ellipse is 16x 2 + y 2 = 16
Dividing by 16 we get x2/1 + y2/ 16 = 1
Major axis is along y-axis a 2 = 16,
=> a = 4, b 2 = 1,
=> b = 1
and c2 = a– b2 = 16 – 1 = 15, ∴ c = 15
foci are (0, ± c) i.e. (0, ± 15)
vertices are (0, ± a) i.e. (0, ±4)
Length of major axis = 2a = 2 × 4 = 8;
Length of minor axis = 2b = 2 × 1 = 2
Eccentricity = e = c/a = 15/4 ;
Length of the 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

The given equation of the ellipse can be written in standard form as
4x2/36 + 9y2/36 = 36/36 => x2/9 + y2/4 = 1
Since the denominator of x 2 > denominator of y 2, the major axis is along the x-axis. Comparing the given equation with the standard equation
x2/a2 + y2/b2 = 1
We have a = 3 and b = 2.
Also, c = √a2 - b2 = √9 - 4 = √5
and e = c/a = √5/3
Hence, the foci, are (√5, 0) and (– √5, 0). vertices, are (3, 0) and (–3, 0); the length of the major axis, 2a is 6 units, the length of the minor axis, 2b is 4 units and the eccentricity of the ellipse, e is √5/ 3
The length of latus rectum
= 2b2/a = 2(22)/3 = 8/3 units.

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

Vertices (± 5, 0), Foci (± 4, 0)
=> (± a, 0) = (±5, 0) and (± ae, 0) = (± 4, 0)
∴ a = 5 and ae = 4
=> e = 4/a = 4/5
Also b 2 = a 2 (1 – e 2) (gives)
b2 = 25(1 – 16/25) = 9
b = 3
∴ The equation of ellipse x2/a2 + y2/b2 = 1 becomes
x2/25 + y2/9 = 1 => 9x2 + 25y2 = 225 which is the equation of the required ellipse.

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

Foci (0, ± 5), vertices (0, ± 13)
(0, ± ae) = (0, ± 5) and (0, ± a) = (0, ± 13)
=> ae = 5 and a = 13 ∴ ae/a = 5/13
b2 = a2 – a2e2 = 132 – 52 = 169 – 25 = 144
∴ b= 12
∴ equation of the required ellipse
= x2/144 + y2/169 = 1

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

Since the vertices are on x-axis, the equation will be of the form.
x2/a2 + y2/b2 = 1
Where a is the semi-major axis.
Given that a = 6, c = ± 4
Therefore from the relation
c2 = a2 – b2 , we get
16 = 36 – b 2
=> b 2 = 20 or b = 2√5
Hence, the equation of the ellipse is
x2/36 + y2/20 = 1

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

Since the vertices are on x-axis, the equation will be of the form
x2/a2 + y2/b2 = 1
Where a is the semi-major axis.
Given a = 3, b = 2
Hence, the equation of the ellipse is
x2/9 + y2/4 = 1

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

Ends of major axis (0, ± √5).
Major axis is the y-axis and a = √5
Ends of minor axis are (± 1, 0)
b = 1
Equation of ellipse is x2/1 + y2/5 = 1

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

Length of major axis = 2a = 26 ∴ a = 13
Foci are (± 5, 0), c = 5,
∴b 2 = a 2 – c 2
= 169 – 25 = 144
Major axis is x-axis.
Equation of ellipse is x2/169 + y2/144 = 1

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

Since the foci are on the y-axis, the major axis will be on y-axis. The equation will be of the form
x2/b2 + y2/a2 = 1
Given 2b = 16 or b = 8 and c = 16
We know that c 2 = a 2 – b 2 or 36 = a 2 – 64
or a2 = 100 or a = 10
Hence, the equation of the ellipse is
x2/64 + y2/100 = 1

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

Since the foci are on the x-axis, the major axis will be on x-axis. The equation will be of the form
x2/a2 + y2/b2 = 1
Given c = ± 3; a = 4
Therefore, c 2 = a 2 – b 2 => 9 = 16 – b 2
and b 2 = 7 => b = √7
Hence, 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 x -axis.

Since the foci are on the x-axis, the equation will be of the form
x2/a2 + y2/b2 = 1
Given that b = 3; c = 4
Therefore, c 2 = a 2 – b 2
or 16 = a2 – 9 => a 2 = 25 => a = 5
Hence, the equation of the ellipse is
x2/25 + y2/9 = 1

19. Find the equation of the ellipse whose centre at (0, 0), major axis on the y-axis and passes through the points (3, 2) and (1, 6)

Major axis is y-axis
Let the ellipse be x2/b2 + y2/a2 = 1
since (3, 2) and (1, 6) lies on it
∴ 9/b+ 4/a2 = 1 …..(i);
1/b2 + 36/a2 = 1 ……(ii)
Subtracting (9 – 1)/b2 + (4 – 36)/a2 = 0
=> 8/b2 – 32/a2 = 0 => 8a2 = 32b2 ∴ a 2 = 4b 2
Putting the value in (i), it become 9/b2 + 4/4b2 = 1
∴ 10/b2 = 1
∴ b 2 = 10
Now, a 2 = 4b 2 = 4 × 10 = 40
∴ Equation of the ellipse x2/10 + y2/40 = 1

20. If major axis on the x-axis and passes through the points (4, 3) and (6, 2), then find the equation for the ellipse that satisfies the given condition.