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JEE Main Math Class 11 Coordinate Geometry Part 5 Questions
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© examsnet.com
Question : 15
Total: 100
The sum of squares of all possible values of
k
, for which area of the region bounded by the parabolas
2
y
2
=
kx
and
ky
2
=
2
(
y
−
x
)
is maximum, is equal to :
[1-Feb-2024 Shift 2]
Your Answer:
Validate
Solution:
👈: Video Solution
k
y
2
=
2
(
y
−
x
)
2
y
2
=
k
x
Point of intersection
→
k
y
2
=
(
y
−
2
y
2
k
)
y
=
0
k
y
=
2
(
1
−
2
y
k
)
k
y
+
4
y
k
=
2
y
=
2
k
+
4
k
=
2
k
k
2
+
4
A
=
2
k
k
2
+
4
∫
0
(
(
y
−
k
y
2
2
)
−
(
2
y
2
k
)
)
⋅
d
y
=
y
2
2
−
(
k
2
+
2
k
)
⋅
y
3
3
|
0
2
k
k
2
+
4
=
(
2
k
k
2
+
4
)
2
[
1
2
−
k
2
+
4
2
k
×
1
3
×
2
k
k
2
+
4
]
=
1
6
×
4
×
(
1
k
+
4
k
)
2
A
⋅
M
≥
G
⋅
M
(
k
+
4
k
)
2
≥
2
k
+
4
k
≥
4
Area is maximum when
k
=
4
k
k
=
2
,
−
2
© examsnet.com
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