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CGPET 2012 Solved Paper
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Section:
Mathematics
© examsiri.com
Question :
145 of 150
Marks:
+1
,
-0
If
e
1
and
e
2
are the eccentricities of the hyperbolas
‌
x
2
a
2
−
‌
y
2
b
2
=
1
and
‌
x
2
a
2
−
‌
y
2
b
2
=
−
1
, then the value of
‌
1
e
1
2
+
‌
1
e
2
2
is
3
2
1
‌
1
2
Validate
Solution:
‌
x
2
a
2
−
‌
y
2
b
2
=
1
⇒
‌
‌
e
1
=
√
1
+
‌
b
2
a
2
=
√
‌
a
2
+
b
2
a
2
and
‌
‌
‌
x
2
a
2
−
‌
y
2
b
2
=
−
1
⇒
‌
‌
‌
y
2
b
2
−
‌
x
2
a
2
=
1
⇒
‌
‌
e
2
=
√
1
+
‌
a
2
b
2
=
√
‌
a
2
+
b
2
b
2
∴
‌
‌
‌
1
e
1
2
+
‌
1
e
2
2
=
‌
a
2
a
2
+
b
2
+
‌
b
2
a
2
+
b
2
⇒
‌
‌
=
1
© examsiri.com
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