Which of the following statements are correct

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Area bounded by ${(y - \sin^{- 1} x)}^{2} = x - x^{2}$ is $\frac{π}{4}$

Area bounded by $x^{2} y^{2} = a^{2} (y^{2} - x^{2})$ and $x = \pm a$ is $4 a^{2}$

Area bounded by $y^{2} = x^{2} (a^{2} - x^{2}) i s \frac{4}{3} a^{2}$

Area bounded between $x = y^{2}$ and $x = 3 - 2 y^{2}$ is 4

answer is A, B, D.

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Detailed Solution

A) $y - \sin^{- 1} x = \pm \sqrt{x - x^{2}}$
$y = \sin^{- 1} x + \sqrt{x - x^{2}}, x - x^{2} \geq 0$
$y = \sin x - \sqrt{x - x^{2}}, 0 \leq x \leq 1$
Question Image
Required are $= \int_{0}^{1} 2 \sqrt{x - x^{2}} d x$
$x = \sin^{2} θ$
$2 \int_{0}^{π / 2} \sin θ . \cos θ .2 \sin θ \cos θ d θ = 4 \int_{0}^{π / 2} \sin^{2} θ . \cos^{2} θ d θ$
$4 \times \frac{1 \times 1}{4 \times 2} \times \frac{π}{2} = \frac{π}{4}$
B) $y^{2} (a^{2} - x^{2}) = a^{2} x^{2}$
Question Image
$y^{2} = \frac{a^{2} x^{2}}{(a^{2} - x^{2})}$
$a^{2} - x^{2} > 0 \Rightarrow - a < x < a$
Required area $= 4 \int_{0}^{a} \frac{a x}{\sqrt{a^{2} - x^{2}}} d x$
$= 2 a \int_{0}^{a} \frac{- 2 x}{\sqrt{a^{2} - x^{2}}} d x$
$= 2 a . {(2 \sqrt{a^{2} - x^{2}})}_{0}^{a}$
$= - 2 a (- a) = 4 a^{2}$
Similarly (B) and (C) are true