NCERT Solutions Class 11th Maths Chapter – 11 Conic Sections Exercise 11.3

NCERT Solutions Class 11th Maths Chapter – 11 Conic Sections Exercise 11.3

NCERT Solutions Class 11th Maths Chapter – 11 Conic Sections Exercise 11.3

?Chapter – 11?

✍Conic Sections✍

?Exercise 11.3?

In each of the Exercises 1 to 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.

1. x²/36 + y²/16 = 1

Solution: Given:
The equation is x²/36 + y²/16 = 1
Here, the denominator of x²/36 is greater than the denominator of y²/16.
So, 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 get
a = 6 and b = 4.
c = √(a² – b²)
= √(36-16)
= √20
= 2√5
Then,
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 = 2 (6) = 12
Length of minor axis = 2b = 2 (4) = 8
Eccentricity, e = c/a = 2√5/6 = √5/3
Length of latus rectum = 2b²/a = (2×16)/6 = 16/3

2. x²/4 + y²/25 = 1

Solution: Given:
The equation is x²/4 + y²/25 = 1
Here, the denominator of y²/25 is greater than the denominator of x²/4.
So, 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 get
a = 5 and b = 2.
c = √(a2 – b2)
= √(25-4)
= √21
Then,
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 = 2 (5) = 10
Length of minor axis = 2b = 2 (2) = 4
Eccentricity, e = c/a = √21/5
Length of latus rectum = 2b²/a = (2×22)/5 = (2×4)/5 = 8/5

3. x²/16 + y²/9 = 1

Solution: Given:
The equation is x²/16 + y²/9 = 1 or x²/42 + y²/32 = 1
Here, the denominator of x²/16 is greater than the denominator of y²/9.
So, 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 get
a = 4 and b = 3.
c = √(a² – b²)
= √(16-9)
= √7
Then,
The coordinates of the foci are (√7, 0) and (-√7, 0).
The coordinates of the vertices are (4, 0) and (-4, 0)
Length of major axis = 2a = 2 (4) = 8
Length of minor axis = 2b = 2 (3) = 6
Eccentricity, e = c/a = √7/4
Length of latus rectum = 2b²/a = (2×32)/4 = (2×9)/4 = 18/4 = 9/2

4. x²/25 + y²/100 = 1

Solution: Given:
The equation is x²/25 + y²/100 = 1
Here, the denominator of y²/100 is greater than the denominator of x²/25.
So, 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 get
b = 5 and a =10.
c = √(a² – b²)
= √(100-25)
= √75
= 5√3
Then,
The coordinates of the foci are (0, 5√3) and (0, -5√3).
The coordinates of the vertices are (0, √10) and (0, -√10)
Length of major axis = 2a = 2 (10) = 20
Length of minor axis = 2b = 2 (5) = 10
Eccentricity, e = c/a = 5√3/10 = √3/2
Length of latus rectum = 2b²/a = (2×52)/10 = (2×25)/10 = 5

5. x²/49 + y²/36 = 1

Solution: Given:
The equation is x²/49 + y²/36 = 1
Here, the denominator of x²/49 is greater than the denominator of y²/36.
So, 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 get
b = 6 and a =7
c = √(a² – b²)
= √(49-36)
= √13
Then,
The coordinates of the foci are (√13, 0) and (-√3, 0).
The coordinates of the vertices are (7, 0) and (-7, 0)
Length of major axis = 2a = 2 (7) = 14
Length of minor axis = 2b = 2 (6) = 12
Eccentricity, e = c/a = √13/7
Length of latus rectum = 2b²/a = (2×62)/7 = (2×36)/7 = 72/7

6. x²/100 + y²/400 = 1

Solution: Given:
The equation is x²/100 + y²/400 = 1
Here, the denominator of y²/400 is greater than the denominator of x²/100.
So, 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 get
b = 10 and a =20.
c = √(a2 – b2)
= √(400-100)
= √300
= 10√3
Then,
The coordinates of the foci are (0, 10√3) and (0, -10√3).
The coordinates of the vertices are (0, 20) and (0, -20)
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 = (2×102)/20 = (2×100)/20 = 10

7. 36×2 + 4y2 = 144

Solution: Given:
The equation is 36x² + 4y² = 144 or x²/4 + y²/36 = 1 or x²/22 + y²/62 = 1
Here, the denominator of y²/62 is greater than the denominator of x²/22.
So, 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 get
b = 2 and a = 6.
c = √(a² – b²)
= √(36-4)
= √32
= 4√2
Then,
The coordinates of the foci are (0, 4√2) and (0, -4√2).
The coordinates of the vertices are (0, 6) and (0, -6)
Length of major axis = 2a = 2 (6) = 12
Length of minor axis = 2b = 2 (2) = 4
Eccentricity, e = c/a = 4√2/6 = 2√2/3
Length of latus rectum = 2b²/a = (2×22)/6 = (2×4)/6 = 4/3

8. 16x² + y² = 16

Solution: Given:
The equation is 16×2 + y² = 16 or x²/1 + y²/16 = 1 or x²/12 + y²/42 = 1
Here, the denominator of y²/42 is greater than the denominator of x²/12.
So, 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 get
b =1 and a =4.
c = √(a2 – b2)
= √(16-1)
= √15
Then,
The coordinates of the foci are (0, √15) and (0, -√15).
The coordinates of the vertices are (0, 4) and (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 latus rectum = 2b²/a = (2×12)/4 = 2/4 = ½

9. 4x² + 9y² = 36

Solution: Given:
The equation is 4x² + 9y² = 36 or x²/9 + y²/4 = 1 or x²/32 + y²/22 = 1
Here, the denominator of x2/32 is greater than the denominator of y²/22.
So, 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 get
a =3 and b =2.
c = √(a² – b²)
= √(9-4)
= √5
Then,
The coordinates of the foci are (√5, 0) and (-√5, 0).
The coordinates of the vertices are (3, 0) and (-3, 0)
Length of major axis = 2a = 2 (3) = 6
Length of minor axis = 2b = 2 (2) = 4
Eccentricity, e = c/a = √5/3
Length of latus rectum = 2b²/a = (2×22)/3 = (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:

10. Vertices (± 5, 0), foci (± 4, 0)

Solution: Given:
Vertices (± 5, 0) and foci (± 4, 0)
Here, the vertices are on the x-axis.
So, the equation of the ellipse will be of the form x²/a² + y²/b² = 1, where ‘a’ is the semi-major axis.
Then, a = 5 and c = 4.
It is known that a² = b² + c²
So, 52 = b² + 42
25 = b² + 16
b² = 25 – 16
b = √9
= 3
∴ The equation of the ellipse is x²/52 + y²/32 = 1 or x²/25 + y²/9 = 1

11. Vertices (0, ± 13), foci (0, ± 5)

Solution: Given:
Vertices (0, ± 13) and foci (0, ± 5)
Here, the vertices are on the y-axis.
So, the equation of the ellipse will be of the form x²/b² + y²/a² = 1, where ‘a’ is the semi-major axis.
Then, a =13 and c = 5.
It is known that a² = b² + c²
132 = b²+52
169 = b² + 15
b² = 169 – 125
b = √144
= 12
∴ The equation of the ellipse is x²/122 + y²/132 = 1 or x²/144 + y²/169 = 1

12. Vertices (± 6, 0), foci (± 4, 0)

Solution: Given:
Vertices (± 6, 0) and foci (± 4, 0)
Here, the vertices are on the x-axis.
So, the equation of the ellipse will be of the form x²/a² + y²/b² = 1, where ‘a’ is the semi-major axis.
Then, a = 6 and c = 4.
It is known that a² = b² + c²
62 = b²+42
36 = b² + 16
b² = 36 – 16
b = √20
∴ The equation of the ellipse is x²/62 + y²/(√20)² = 1 or x²/36 + y²/20 = 1

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

Solution: Given:
Ends of major axis (± 3, 0) and ends of minor axis (0, ±2)
Here, the major axis is along the x-axis.
So, the equation of the ellipse will be of the form x²/a² + y²/b² = 1, where ‘a’ is the semi-major axis.
Then, a = 3 and b = 2.
∴ The equation for the ellipse x²/32 + y²/22 = 1 or x²/9 + y²/4 = 1

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

Solution: Given:
Ends of major axis (0, ±√5) and ends of minor axis (±1, 0)
Here, the major axis is along the y-axis.
So, the equation of the ellipse will be of the form x²/b² + y²/a² = 1, where ‘a’ is the semi-major axis.
Then, a = √5 and b = 1.
∴ The equation for the ellipse x²/12 + y²/(√5)² = 1 or x2/1 + y²/5 = 1

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

Solution: Given:
Length of major axis is 26 and foci (±5, 0)
Since the foci are on the x-axis, the major axis is along the x-axis.
So, the equation of the ellipse will be of the form x²/a² + y²/b² = 1, where ‘a’ is the semi-major axis.
Then, 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
∴ The equation of the ellipse is x²/132 + y²/122 = 1 or x²/169 + y²/144 = 1

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

Solution: Given:
Length of minor axis is 16 and foci (0, ±6).
Since the foci are on the y-axis, the major axis is along the y-axis.
So, the equation of the ellipse will be of the form x²/b² + y²/a² = 1, where ‘a’ is the semi-major axis.
Then, 2b =16
b = 8 and c = 6.
It is known that a² = b² + c²
a² = 82 + 62
= 64 + 36
=100
a = √100
= 10
∴ The equation of the ellipse is x²/82 + y²/102 =1 or x²/64 + y²/100 = 1

17. Foci (±3, 0), a = 4

Solution: Given:
Foci (±3, 0) and a = 4
Since the foci are on the x-axis, the major axis is along the x-axis.
So, the equation of the ellipse will be of the form x²/a² + y²/b² = 1, where ‘a’ is the semi-major axis.
Then, c = 3 and a = 4.
It is known that a² = b² + c².
a² = 82 + 62
= 64 + 36
= 100
16 = b² + 9
b² = 16 – 9
= 7
∴ The equation of the ellipse is x²/16 + y²/7 = 1

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

Solution: Given:
b = 3, c = 4, centre at the origin and foci on the x axis.
Since the foci are on the x-axis, the major axis is along the x-axis.
So, the equation of the ellipse will be of the form x²/a² + y²/b² = 1, where ‘a’ is the semi-major axis.
Then, b = 3 and c = 4.
It is known that a² = b² + c²
a² = 32 + 42
= 9 + 16
=25
a = √25
= 5
∴ The equation of the ellipse is x²/52 + y²/32 or x²/25 + y²/9 = 1

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

Solution: Given:
Centre at (0, 0), major axis on the y-axis and passes through the points (3, 2) and (1, 6).
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 x²/b² + y²/a² = 1, where ‘a’ is the semi-major axis.
The ellipse passes through points (3, 2) and (1, 6).
So, by putting the values x = 3 and y = 2, we get,
32/b² + 22/a² = 1
9/b² + 4/a²…. (1)
And by putting the values x = 1 and y = 6, we get,
11/b² + 62/a² = 1
1/b² + 36/a² = 1 …. (2)
On solving equation (1) and (2), we get
b² = 10 and a² = 40.
∴ The equation of the ellipse is x²/10 + y²/40 = 1 or 4×2 + y² = 40

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

Solution: Given:
Major axis on the x-axis and passes through the points (4, 3) and (6, 2).
Since the major axis is on the x-axis, the equation of the ellipse will be the form
x²/a² + y²/b² = 1…. (1) [Where ‘a’ is the semi-major axis.]
The ellipse passes through points (4, 3) and (6, 2).
So by putting the values x = 4 and y = 3 in equation (1), we get,
16/a² + 9/b² = 1 …. (2)
Putting, x = 6 and y = 2 in equation (1), we get,
36/a² + 4/b² = 1 …. (3)
From equation (2)
16/a² = 1 – 9/b²
1/a² = (1/16 (1 – 9/b²)) …. (4)
Substituting the value of 1/a² in equation (3) we get,
36/a² + 4/b² = 1
36(1/a²) + 4/b² = 1
36[1/16 (1 – 9/b²)] + 4/b² = 1
36/16 (1 – 9/b²) + 4/b² = 1
9/4 (1 – 9/b²) + 4/b² = 1
9/4 – 81/4b² + 4/b² = 1
-81/4b² + 4/b² = 1 – 9/4
(-81+16)/4b² = (4-9)/4
-65/4b² = -5/4
-5/4(13/b²) = -5/4
13/b² = 1
1/b² = 1/13
b² = 13
Now substitute the value of b² in equation (4) we get,
1/a² = 1/16(1 – 9/b²)
= 1/16(1 – 9/13)
= 1/16((13-9)/13)
= 1/16(4/13)
= 1/52
a² = 52
Equation of ellipse is x²/a² + y2/b² = 1
By substituting the values of a² and b² in above equation we get,
x²/52 + y²/13 = 1