13. Find the area bounded by the curves $x ^ { 2 } = y , x ^ { 2 } = - y$ and $y ^ { 2 } = 4 x - 3$.

Let $S$ denote the locus of the mid-points of those chords of the parabola $y ^ { 2 } = x$, such that the area of the region enclosed between the parabola and the chord is $\frac { 4 } { 3 }$. Let $\mathcal { R }$ denote the region lying in the first quadrant, enclosed by the parabola $y ^ { 2 } = x$, the curve $S$, and the lines $x = 1$ and $x = 4$.
Then which of the following statements is (are) TRUE?

(A)	$( 4 , \sqrt { 3 } ) \in S$
(B)	$( 5 , \sqrt { 2 } ) \in S$
(C)	Area of $\mathcal { R }$ is $\frac { 14 } { 3 } - 2 \sqrt { 3 }$
(D)	Area of $\mathcal { R }$ is $\frac { 14 } { 3 } - \sqrt { 3 }$

The area bounded between the parabolas $x^{2} = \frac{y}{4}$ and $x^{2} = 9y$, and the straight line $y = 2$ is
(1) $20\sqrt{2}$
(2) $\frac{10\sqrt{2}}{3}$
(3) $\frac{20\sqrt{2}}{3}$
(4) $10\sqrt{2}$

The area (in sq. units) of the region $\{(x, y) : y^2 \geq 2x$ and $x^2 + y^2 \leq 4x, x \geq 0, y \geq 0\}$ is:
(1) $\pi - \frac{4\sqrt{2}}{3}$
(2) $\pi - \frac{8}{3}$
(3) $\pi - \frac{4}{3}$
(4) $\frac{\pi}{2} - \frac{2\sqrt{2}}{3}$

The area (in sq. units) of the region $\{(x,y): y^2 \geq 2x$ and $x^2 + y^2 \leq 4x, x \geq 0, y \geq 0\}$ is: (1) $\pi - \frac{4\sqrt{2}}{3}$ (2) $\pi - \frac{8}{3}$ (3) $\pi - \frac{4}{3}$ (4) $\frac{\pi}{2} - \frac{2\sqrt{2}}{3}$

The area of the region (in sq. units), enclosed by the circle $x ^ { 2 } + y ^ { 2 } = 2$ which is not common to the region bounded by the parabola $y ^ { 2 } = x$ and the straight line $y = x$, is
(1) $\frac { 1 } { 6 } (24 \pi - 1)$
(2) $\frac { 1 } { 3 } (6 \pi - 1)$
(3) $\frac { 1 } { 3 } (12 \pi - 1)$
(4) $\frac { 1 } { 6 } (12 \pi - 1)$

Area (in sq. units) of the region outside $\frac{|x|}{2} + \frac{|y|}{3} = 1$ and inside the ellipse $\frac{x^{2}}{4} + \frac{y^{2}}{9} = 1$ is
(1) $6(\pi - 2)$
(2) $3(\pi - 2)$
(3) $3(4 - \pi)$
(4) $6(4 - \pi)$

The area of the region bounded by $y ^ { 2 } = 8 x$ and $y ^ { 2 } = 16 ( 3 - x )$ is equal to
(1) $\frac { 32 } { 3 }$
(2) $\frac { 40 } { 3 }$
(3) 16
(4) 9

If the tangent at a point P on the parabola $\mathrm { y } ^ { 2 } = 3 \mathrm { x }$ is parallel to the line $x + 2 y = 1$ and the tangents at the points $Q$ and $R$ on the ellipse $\frac { x ^ { 2 } } { 4 } + \frac { y ^ { 2 } } { 1 } = 1$ are perpendicular to the line $x - y = 2$, then the area of the triangle $P Q R$ is: (1) $\frac { 9 } { \sqrt { 5 } }$ (2) $5 \sqrt { 3 }$ (3) $\frac { 3 } { 2 } \sqrt { 5 }$ (4) $3 \sqrt { 5 }$

The area of the region, inside the circle $( x - 2 \sqrt { 3 } ) ^ { 2 } + y ^ { 2 } = 12$ and outside the parabola $y ^ { 2 } = 2 \sqrt { 3 } x$ is:
(1) $3 \pi + 8$
(2) $6 \pi - 16$
(3) $3 \pi - 8$
(4) $6 \pi - 8$

Let the area enclosed between the curves $| y | = 1 - x ^ { 2 }$ and $x ^ { 2 } + y ^ { 2 } = 1$ be $\alpha$. If $9 \alpha = \beta \pi + \gamma ; \beta , \gamma$ are integers, then the value of $| \beta - \gamma |$ equals.
(1) 27
(2) 33
(3) 15
(4) 18

Let $a , b$ be real numbers, and let $O$ be the origin of the coordinate plane. It is known that the graph of the quadratic function $f ( x ) = a x ^ { 2 }$ and the circle $\Omega : x ^ { 2 } + y ^ { 2 } - 3 y + b = 0$ both pass through point $P \left( 1 , \frac { 1 } { 2 } \right)$, and let point $C$ be the center of $\Omega$.
Find the area of the region bounded by the graph of $y = f ( x )$ above and the lower semicircular arc of $\Omega$.