4x(x² + y²) – 8(x² + y2)
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Answer:
Because x
2
+y
2
=9;y
2
=8x
So x
2
+8x=9
x
2
+8x–9=0
⟹(x+9)(x−1)=0
⟹x=1,x=−9
Therefore y=±2
2
Point of intersections are p(1,2
2
,1,−2
2
)
y
1
=
8x
=2
2
x
1/2
y
2
=
9−x
2
Area = Area OPAQO = 2 Area OPAMO
Area = 2 (Area OPMO + Area APMA)
=2[∫
0
1
y
1
dx+∫
1
3
y
2
dx]
=2[∫
0
1
2
2
x
1/2
dx+∫
1
3
3
2
–x
2
dx]
=2[
3/2
2
2
(x
3/2
)
0
1
+(
2
x
3
2
–x
2
+
2
9
sin
−1
3
x
)
1
3
]
=2[
2
4
2
.1+
2
9
sin
−1
1−
2
1
×
8
−
2
9
sin
−1
3
1
]
=2[
3
2
+
4
9π
−
2
9
sin
−1
3
1
] sq. Unit
Explanation:
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