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AP EAMCET Engineering 2018 Apr 22 Shift 1 Paper
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© examsnet.com
Question : 57
Total: 160
An ellipse having the coordinate axes as its axes and its major axis along Y-axis, passes through the point (-3 , 1) and has eccentricity
√
2
5
. Then its equation is
3
x
2
+
5
y
2
- 15 = 0
5
x
2
+
3
y
2
- 32 = 0
3
x
2
+
5
y
2
- 32 = 0
5
x
2
+
3
y
2
- 48 = 0
Validate
Solution:
The equation of the ellipse is given as.
x
2
a
2
+
y
2
b
2
=
1
and,
e
=
√
2
5
∴
√
a
2
−
b
2
a
2
=
√
2
5
a
2
−
b
2
a
2
=
2
5
5
a
2
−
5
b
2
=
2
a
2
a
2
=
5
b
2
3
The ellipse passes through the point (-3,1) .
(
−
3
)
2
a
2
+
(
1
)
2
b
2
=
1
9
h
2
+
5
b
2
3
=
5
b
4
3
b
2
=
32
5
Substituting the value of
a
and
b
in general equation we get.
x
2
(
32
3
)
+
y
2
(
35
5
)
=
1
3
x
2
+
5
y
2
−
32
=
0
© examsnet.com
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