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JEE Main 26 June 2022 Shift 1 Solved Paper
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© examsnet.com
Question : 23
Total: 90
Let the common tangents to the curves
4
(
x
2
+
y
2
)
=
9
and
y
2
=
4
x
intersect at the point Q. Let an ellipse, centered at the origin 0 , has lengths of semi-minor and semi-major axes equal to
O
Q
and 6, respectively. If e and I respectively denote the eccentricity and the length of the latus rectum of this ellipse, then
l
e
2
is equal to____
[26-Jun-2022-Shift-1]
Your Answer:
Validate
Solution:
👈: Video Solution
Let
y
=
m
x
+
c
is the common tangent
So
c
=
1
m
=
±
3
2
√
1
+
m
2
⇒
m
2
=
1
3
So equation of common tangents will be
y
=
±
1
√
3
x
±
√
3
, which intersects at
Q
(
−
3
,
0
)
Major axis and minor axis of ellipse are 12 and 6.
So eccentricity
e
2
=
1
−
1
4
=
3
4
and length of latus rectum
=
2
b
2
a
=
3
Hence,
l
e
2
=
3
3
∕
4
=
4
© examsnet.com
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