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JEE Main Math Class 11 Coordinate Geometry Part 1 Questions
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© examsnet.com
Question : 68
Total: 100
The ellipse
x
2
+
4
y
2
=
4
is inscribed in a rectangle aligned with the coordinate axes, which in turn is inscribed in another ellipse that passes through the point (4,0) . Then the equation of the ellipse is :
[2009]
x
2
+
12
y
2
=
16
4
x
2
+
48
y
2
=
48
4
x
2
+
64
y
2
=
48
x
2
+
16
y
2
=
16
Validate
Solution:
The given equation of ellipse is
x
2
4
+
y
2
1
=
1
So,
A
=
(
2
,
0
)
and
B
=
(
0
,
1
)
If
P
Q
R
S
is the rectangle in which it is inscribed, then
P
=
(
2
,
1
)
Let
x
2
a
2
+
y
2
b
2
=
1
be the ellipse
circumscribing the rectangle
P
Q
R
S
.
Then it passed through
P
(
2
,
1
)
∴
4
a
2
+
1
b
2
=
1
. . . . . (i)
Also, given that, it passes through (4,0)
∴
16
a
2
+
0
=
1
⇒
a
2
=
16
⇒
b
2
=
4
∕
3
[putting
a
2
=
16
in eq
n
(
i
)
]
∴
The required equation of ellipse is
x
2
16
+
y
2
4
∕
3
=
1
or
x
2
+
12
y
2
=
16
© examsnet.com
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