Chapter 11 Conic Sections Ex 11.3 Solutions
Question - 11 : - Find the equation for the ellipse that satisfies the given conditions:Vertices(0, ± 13), foci (0, ± 5)
Answer - 11 : -
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
∴ Theequation of the ellipse is x2/122 + y2/132 =1 or x2/144 + y2/169 = 1
Question - 12 : - Find the equation for the ellipse that satisfies the given conditions:Vertices (± 6, 0), foci (± 4, 0)
Answer - 12 : -
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
∴ Theequation of the ellipse is x2/62 + y2/(√20)2 =1 or x2/36 + y2/20 = 1
Question - 13 : - Find the equation for the ellipse that satisfies the given conditions:Ends of major axis (0, ±√5), ends of minor axis (±1, 0)
Answer - 13 : -
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.
∴ Theequation for the ellipse x2/12 + y2/(√5)2 =1 or x2/1 + y2/5 = 1
Question - 14 : - Find the equation for the ellipse that satisfies the given conditions:Length of major axis 26, foci (±5, 0)
Answer - 14 : -
Given:
Length of major axis is 26 and foci (±5, 0)
Since the foci are on the x-axis, the major axis is along thex-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 equationof the ellipse is x2/132 + y2/122 =1 or x2/169 + y2/144 = 1
Question - 15 : - Find the equation for the ellipse that satisfies the given conditions:Length of minor axis 16, foci (0, ±6).
Answer - 15 : -
Given:
Length of minor axis is 16 and foci (0, ±6).
Since the foci are on the y-axis, the major axis is along they-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
∴ Theequation of the ellipse is x2/82 + y2/102 =1or x2/64 + y2/100 = 1
Question - 16 : - Find the equation for the ellipse that satisfies the given conditions:Foci (±3, 0), a = 4
Answer - 16 : -
Given:
Foci (±3, 0) and a = 4
Since the foci are on the x-axis, the major axis is along thex-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
∴ Theequation of the ellipse is x2/16 + y2/7 = 1
Question - 17 : - 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 - 17 : -
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 thex-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
∴ Theequation of the ellipse is x2/52 + y2/32 orx2/25 + y2/9 = 1
Question - 18 : - 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 - 18 : -
Given:
Centre at (0, 0), major axis on the y-axis and passes throughthe 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.
∴ Theequation of the ellipse is x2/10 + y2/40 = 1 or 4x2 +y 2 = 40
Question - 19 : - 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 - 19 : -
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 theellipse 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 inabove equation we get,
x2/52 + y2/13 = 1