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JEE Main Math Class 11 Coordinate Geometry Part 1 Questions
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© examsnet.com
Question : 24
Total: 100
A chord is drawn through the focus of the parabola
y
2
=
6
x
such that its distance from the vertex of this parabola is
√
5
2
,
then its slope can be:
[Online April 19, 2014]
√
5
2
√
3
2
2
√
5
2
√
3
Validate
Solution:
Equation of parabola,
y
2
=
6
x
⇒
y
2
=
4
×
3
2
x
∴
Focus
=
(
3
2
,
0
)
Let equation of chord passing through focus be
a
x
+
b
y
+
c
=
0
.
.
.
(
1
)
Since chord is passing through
(
3
2
,
0
)
∴
Put
x
=
3
2
,
y
=
0
in eqn
(
1
)
,
we get
3
2
a
+
c
=
0
⇒
c
=
−
3
2
a
.
.
.
(
2
)
distance of chord from origin is
√
5
2
√
5
2
=
|
a
(
0
)
+
b
(
0
)
+
c
√
a
2
+
b
2
|
=
c
√
a
2
+
b
2
5
4
=
c
2
a
2
+
b
2
⇒
a
2
+
b
2
=
4
5
c
2
Putting value of
c
from
(
2
)
,
we get
a
2
+
b
2
=
4
5
×
9
4
a
2
b
2
=
9
5
a
2
−
a
2
=
4
5
a
2
a
2
b
2
=
5
4
,
a
b
=
±
√
5
2
Slope of chord,
d
y
d
x
=
−
a
b
=
−
(
±
√
5
2
)
=
∓
√
5
2
© examsnet.com
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