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Byju's Answer
Standard IX
History
Aurangzeb
Find the area...
Question
Find the area bounded by the ellipse
x
2
a
2
+
y
2
b
2
=
1
and the ordinates
x
=
0
and
x
=
a
e
, where
b
2
=
a
2
(
1
−
e
2
)
and
e
<
1
.
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Solution
Given line
y
=
x
and the circle
x
2
+
y
2
=
32
⇒
x
2
+
y
2
=
16
×
2
=
4
2
×
2
⇒
x
2
+
y
2
=
(
4
√
2
)
2
Centre
=
(
0
,
0
)
Radius
=
4
√
2
Drawing the diagram
⇒
x
2
+
x
2
=
32
⇒
2
x
2
=
32
⇒
x
2
=
16
∴
x
=
±
14
When
x
=
4
y
=
x
=
4
When
x
=
−
4
y
=
x
=
−
4
So, point is
(
4
,
4
)
and
(
−
4
,
−
4
)
As point
M
is in
1
s
t
Quadrant
∴
M
=
(
4
,
4
)
∴
Required Area = Area of OMNO + Area NMAN
=
∫
4
0
y
1
d
x
+
∫
4
√
2
0
y
2
d
x
Where
y
1
=
x
and
y
2
=
√
(
4
√
2
)
2
−
x
2
=
∫
4
0
x
d
x
+
=
∫
4
√
2
4
√
(
4
√
2
)
2
−
x
2
d
x
=
[
x
2
2
]
4
0
+
⎡
⎢ ⎢ ⎢ ⎢
⎣
x
2
√
(
4
√
2
)
2
−
x
2
+
(
4
√
2
)
2
2
sin
−
1
x
4
√
2
⎤
⎥ ⎥ ⎥ ⎥
⎦
4
√
2
4
=
16
2
+
[
32
2
sin
−
1
1
−
(
2
√
32
−
16
+
32
2
sin
−
1
(
1
√
2
)
)
]
=
8
+
16
π
2
−
2
√
16
−
16
π
4
=
8
+
8
π
−
8
−
4
π
=
4
π
Method 2:
Let
∠
A
O
M
=
θ
We know the slope of
y
=
x
is
m
=
1
⇒
tan
θ
=
1
⇒
tan
θ
=
1
⇒
θ
=
π
4
Area of a sector of a circle
=
θ
2
π
×
π
r
2
=
θ
r
2
2
=
π
4
×
(
4
√
2
)
2
2
=
π
×
32
8
=
4
π
Final answer:
Therefore, required area
=
4
π
square units
.
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0
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y
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