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Question
Mathematics
The area included between the parabolas y2=4a (x+a) and y2=4b(x-a), b > a> 0, is
Q. The area included between the parabolas
y
2
=
4
a
(
x
+
a
)
and
y
2
=
4
b
(
x
−
a
)
,
b
>
a
>
0
, is
1730
253
Application of Integrals
Report Error
A
3
4
2
b
2
b
−
a
a
sq. units
100%
B
3
8
8
a
2
b
−
a
b
sq. units
0%
C
3
4
2
a
2
b
−
a
b
sq. units
0%
D
3
8
8
b
2
b
−
a
a
sq. units
0%
Solution:
We have,
y
2
=
4
a
(
x
+
a
)
...
(
i
)
, a parabola with vertex
(
−
a
,
0
)
and
y
2
=
4
b
(
x
−
a
)
...
(
ii
)
, a parabola with vertex
(
a
,
0
)
Solving
(
i
)
and
(
ii
)
, we get
y
=
±
a
b
−
a
8
b
A
=
2
∫
0
b
−
a
a
8
b
(
(
4
b
y
2
+
a
)
−
(
4
a
y
2
−
a
)
)
d
y
=
2
∫
0
b
−
a
a
8
b
(
2
a
−
4
ab
(
b
−
a
)
y
2
)
d
y
=
2
[
2
a
y
−
12
ab
b
−
a
y
3
]
0
b
−
a
a
8
b
=
2
[
2
a
×
b
−
a
a
8
b
−
12
ab
b
−
a
(
b
−
a
a
8
b
)
3
]
=
4
a
2
b
−
a
8
b
−
3
4
a
2
b
−
a
8
b
=
3
8
a
2
b
−
a
8
b
sq. units