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JEE Main Math Class 11 Coordinate Geometry Part 3 Questions
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© examsnet.com
Question : 56
Total: 100
A ray of light through
(
2
,
1
)
is reflected at a point
P
on the
Y
-axis and then passes through the point
(
5
,
3
)
. If this reflected ray is the directrix of an ellipse with eccentricity
1
3
and the distance of the nearer focus from this directrix is
8
√
53
, then the equation of the other directrix can be
[2021, 27 July Shift-1]
11
x
+
7
y
+
8
=
0
or
11
x
+
7
y
−
15
=
0
11
x
−
7
y
−
8
=
0
or
11
x
+
7
y
+
15
=
0
2
x
−
7
y
+
29
=
0
or
2
x
−
7
y
−
7
=
0
2
x
−
7
y
−
39
=
0
or
2
x
−
7
y
−
7
=
0
Validate
Solution:
Equation of the reflected ray will be
L
⇒
y
−
3
x
−
5
=
m
Now, the image of
(
2
,
1
)
w.r.t. line
x
=
0
should lie on the reflected line.
Image of
(
2
,
1
)
=
(
−
2
,
1
)
So,
1
−
3
−
2
−
5
=
m
⇒
m
=
2
7
So, equation of reflected ray
⇒
y
−
3
x
−
5
=
2
7
c
7
y
−
21
=
2
x
−
10
⇒
2
x
−
7
y
+
11
=
0
(equation of directrix)
Now,
e
=
1
3
(given)
So,
a
e
−
a
e
=
a
1
∕
3
−
1
3
a
=
8
a
3
⇒
8
a
3
=
8
√
53
(given)
So,
a
=
3
√
53
and
2
a
e
=
(
2
⋅
3
√
53
)
×
(
3
1
)
=
18
√
53
Now, another directrix will be parallel to the first directrix and lie at a distance of
18
√
53
units.
So, let the equation of another directrix be
2
x
−
7
y
+
λ
=
0
Accordingly,
|
λ
−
11
|
√
2
2
+
7
2
=
18
√
53
⇒
|
λ
−
11
|
=
18
⇒
λ
=
11
±
18
⇒
λ
=
29
or
−
7
So, equation of another directrix will be
2
x
−
7
y
+
29
=
0
or
2
x
−
7
y
−
7
=
0
© examsnet.com
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