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VITEEE 2017 Solved Paper
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© examsnet.com
Question : 114
Total: 125
If the eccentricity of the hyperbola
x
2
−
y
2
c
o
s
e
c
2
α = 25 is
√
5
times the eccentricity of the ellipse
x
2
c
o
s
e
c
2
α
+
y
2
= 5 , then α is equal to :
t
a
n
−
1
√
2
s
i
n
−
1
√
3
4
t
a
n
−
1
√
2
5
s
i
n
−
1
√
2
5
Validate
Solution:
Eccentrcity of
x
2
25
−
y
2
25
s
i
n
2
α
= 1 is
√
1
+
s
i
n
2
α
Eccentricity of
x
2
5
s
i
n
2
α
+
y
2
5
= 1 is
√
1
−
s
i
n
2
α
Given ,
√
1
+
s
i
n
2
α
=
√
5
√
1
−
s
i
n
2
α
⇒
s
i
n
2
α
=
2
3
⇒ α =
s
i
n
−
1
√
2
3
=
t
a
n
−
1
√
2
© examsnet.com
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