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JEE Main Math Class 11 Coordinate Geometry Part 2 Questions
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© examsnet.com
Question : 94
Total: 100
A hyperbola passes through the point
P
(
√
2
,
√
3
)
and has foci at (±2,0) . Then the tangent to this hyperbola at
P
also passes through the point :
[2017]
(
−
√
2
,
−
√
3
)
(
3
√
2
,
2
√
3
)
(
2
√
2
,
3
√
3
)
(
√
3
,
√
2
)
Validate
Solution:
Equation of hyperbola is
x
2
a
2
−
y
2
b
2
=
1
foci is
(
±
2
,
0
)
⇒
a
e
=
±
2
⇒
a
2
e
2
=
4
Since
b
2
=
a
2
(
e
2
−
1
)
b
2
=
a
2
e
2
−
a
2
∴
a
2
+
b
2
=
4
. . . . (i)
Hyperbola passes through
(
√
2
,
√
3
)
∴
2
a
2
−
3
b
2
=
1
. . . . (ii)
2
4
−
b
2
−
3
b
2
=
1
[from (i)]
⇒
b
4
+
b
2
−
12
=
0
⇒
(
b
2
−
3
)
(
b
2
+
4
)
=
0
⇒
b
2
=
3
b
2
=
−
4
(Not possible)
For
b
2
=
3
⇒
a
2
=
1
∴
x
2
1
−
y
2
3
=
1
Equation of tangent is
√
2
x
1
−
√
3
y
3
=
1
Clearly
(
2
√
2
,
3
√
3
)
satisfies it.
© examsnet.com
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