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The area of the circle x2 + y2 = 16 enterior to the parabola y2 = 6x is
The area we want is –
At intersection points, y2 = 6x = 16 – x2
Or, x2 + 6x – 16 = 0
i.e., (x + 8)(x – 2) = 0
i.e., x = -8, 2
Now x = -8 ⇒ y2 = 16 – (-8)2 = 16 – 64 = -48 < 0, which is not possible for y ∈ ℝ
So, x = 2
and y2 = 16 – 22
= 16 – 4
= 12
Or, y = √12 = ± 2√3
So, our bounds are y = -2√3 to y = 2√3
The area A enclosed is –
A =
Since both the functions are symmetrical about the x – axis,
A =
=
(Ans)