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Mathematics
Mathematics
Problem
If the length of latusrectum of ellipse
E
1
:
4
(
x
+
y
−
1
)
2
+
2
(
x
−
y
+
3
)
2
=
8
E_1: 4(x+y-1)^2 + 2(x-y+3)^2 = 8
E
1
:
4
(
x
+
y
−
1
)
2
+
2
(
x
−
y
+
3
)
2
=
8
and
E
2
:
x
2
p
+
y
2
p
2
=
1
,
(
0
<
p
<
1
)
E_2: \frac{x^2}{p} + \frac{y^2}{p^2} = 1, (0 < p < 1)
E
2
:
p
x
2
+
p
2
y
2
=
1
,
(
0
<
p
<
1
)
are equal, then area of ellipse
E
2
E_2
E
2
, is a.
π
2
\frac{\pi}{2}
2
π
b.
π
2
\frac{\pi}{\sqrt{2}}
2
π
c.
π
2
2
\frac{\pi}{2\sqrt{2}}
2
2
π
d.
π
4
\frac{\pi}{4}
4
π
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