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(D) The vertex of the parabola $y^2 = 4ax$ is $(0,0)$. The length of the latus rectum is $4a$. The diameter of the circle is $\frac{3}{4} 	imes 4a =

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$\left(\frac{2+\sqrt{2}}{2}\right)a^2$

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(D) The vertex of the parabola $y^2 = 4ax$ is $(0,0)$. The length of the latus rectum is $4a$. The diameter of the circle is $\frac{3}{4} 	imes 4a = 3a$. Thus,the radius $r = \frac{3a}{2}$. The equation of the circle is $x^2 + y^2 = \left(\frac{3a}{2}ight)^2 = \frac{9a^2}{4}$,or $4(x^2 + y^2) = 9a^2$. Substituting $y^2 = 4ax$ into the circle equation: $4(x^2 + 4ax) = 9a^2$,which simplifies to $4x^2 + 16ax - 9a^2 = 0$. Solving for $x$: $x = \frac{-16a \pm \sqrt{(16a)^2 - 4(4)(-9a^2)}}{2(4)} = \frac{-16a \pm \sqrt{256a^2 + 144a^2}}{8} = \frac{-16a \pm 20a}{8}$. Since $x > 0$,we take $x = \frac{4a}{8} = \frac{a}{2}$. For $x = \frac{a}{2}$,$y^2 = 4a(\frac{a}{2}) = 2a^2$,so $y = \pm \sqrt{2}a$. The points are $P(\frac{a}{2}, \sqrt{2}a)$ and $Q(\frac{a}{2}, -\sqrt{2}a)$. The length of the chord $PQ = 2\sqrt{2}a$. The latus rectum $L_1L_2$ is at $x = a$,with $y^2 = 4a(a) = 4a^2$,so $L_1(a, 2a)$ and $L_2(a, -2a)$. The length $L_1L_2 = 4a$. The area of the trapezium $PL_1L_2Q$ is $\frac{1}{2} 	imes (PQ + L_1L_2) 	imes 	ext{height}$. The height is the distance between $x = a$ and $x = \frac{a}{2}$,which is $a - \frac{a}{2} = \frac{a}{2}$. Area $= \frac{1}{2} 	imes (2\sqrt{2}a + 4a) 	imes \frac{a}{2} = \frac{1}{4} 	imes 2a(\sqrt{2} + 2) = \left(\frac{2+\sqrt{2}}{2}ight)a^2$.

$A$ circle is described whose centre is the vertex and whose diameter is three-quarters of the latus rectum of the parabola $y^2 = 4ax$. If $PQ$ is the common chord of the circle and the parabola and $L_1L_2$ is the latus rectum,then the area of the trapezium $PL_1L_2Q$ is:

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