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EAMCET Engineering 2002 Solved Paper
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© examsnet.com
Question : 133
Total: 200
If
e
and
e
′
are the eccentricities of the ellipse
5
x
2
+
9
y
2
=
45
and the hyperbola
5
x
2
−
4
y
2
=
45
respectively, then
e
e
′
is equal to
1
4
5
9
Validate
Solution:
Equation of ellipse is
5
x
2
+
9
y
2
=
45
or
x
2
9
+
y
2
5
=
1
. . . (i)
e
=
√
1
−
b
2
a
2
=
√
1
−
5
9
=
2
3
and equation of hyperbola is
5
x
2
−
4
y
2
=
45
or
x
2
9
−
y
2
(
45
4
)
=
1
. . . (ii)
e
′
=
√
1
+
b
2
a
2
=
√
1
+
45
36
=
√
81
36
=
9
6
=
3
2
Then,
e
e
′
=
2
3
⋅
3
2
=
1
© examsnet.com
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