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Test Index
TS EAMCET 18-July-2022 Shift 2 Solved Paper
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© examsnet.com
Question : 59
Total: 160
P
(
θ
1
)
and
Q
(
θ
2
)
are two points on the ellipse
x
2
a
2
+
y
2
b
2
=
1
with eccentricity
e
. If
P
S
Q
is a focal chord and
tan
(
θ
1
2
)
tan
(
θ
2
2
)
=
−
(
2
√
2
+
3
)
, then
e
and
S
are
1
√
3
,
(
a
√
3
,
0
)
1
√
3
,
(
−
a
√
3
,
0
)
1
√
2
,
(
a
√
2
,
0
)
1
√
2
,
(
−
a
√
2
,
0
)
Validate
Solution:
👈: Video Solution
If
S
≡
(
a
e
,
0
)
, then
tan
θ
1
2
⋅
tan
θ
2
2
=
e
−
1
e
+
1
If
S
≡
(
−
a
e
,
0
)
, then
tan
θ
1
2
⋅
tan
θ
2
2
=
e
+
1
e
−
1
⇒
|
e
−
1
e
+
1
|
<
1
and
|
e
+
1
e
−
1
|
>
1
According to the question,
tan
θ
1
2
⋅
tan
θ
2
2
=
−
(
2
√
2
+
3
)
⇒
|
tan
θ
1
2
⋅
tan
θ
2
2
|
=
2
√
2
+
3
>
1
⇒
e
+
1
e
−
1
=
−
(
2
√
2
+
3
)
⇒
e
=
1
√
2
In this case,
S
≡
(
−
a
e
,
0
)
i.e.
S
≡
(
−
a
√
2
,
0
)
© examsnet.com
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