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JEE Mains 6-Apr-2023 Shift 2 Solved Paper
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© examsnet.com
Question : 23
Total: 90
Let the eccentricity of an ellipse
x
2
a
2
+
y
2
b
2
=
1
is reciprocal to that of the hyperbola
2
x
2
−
2
y
2
=
1
. If the ellipse intersects the hyperbola at right angles, then square of length of the latus-rectum of the ellipse is _______ :
[6-Apr-2023 shift 2]
Your Answer:
Validate
Solution:
E
:
x
2
a
2
+
y
2
b
2
=
1
→
e
H
:
x
2
−
y
2
=
1
2
⇒
e
′
=
√
2
e
=
1
√
2
∵
e
2
=
1
2
1
−
b
2
a
2
=
1
2
⇒
b
2
a
2
=
1
2
a
2
=
2
b
2
E
&
H
are at right angle
they are confocal
Focus of Hyperbola
=
focus of ellipse
(
±
1
√
2
⋅
√
2
,
0
)
=
(
±
a
√
2
,
0
)
a
=
√
2
∵
a
2
=
2
b
2
⇒
b
2
=
1
Length of LR
=
2
b
2
a
=
2
(
1
)
√
2
=
√
2
Square of LR
=
2
© examsnet.com
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