Let AB be the latus rectum of the parabola y^ | vedclass

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(D) The area of the region $T$ bounded by the parabola $y^2 = 4ax$ and the latus rectum $x = a$ is given by:$	ext{Area}(T) = 2 \int_0^a \sqrt{4ax} \,

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$\frac{1}{\sqrt{3}}$

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(D) The area of the region $T$ bounded by the parabola $y^2 = 4ax$ and the latus rectum $x = a$ is given by:$	ext{Area}(T) = 2 \int_0^a \sqrt{4ax} \, dx = 4\sqrt{a} \int_0^a x^{1/2} \, dx = 4\sqrt{a} \left[ \frac{2}{3} x^{3/2} ight]_0^a = \frac{8}{3} a^2$.Let the coordinates of $R$ be $(x, y)$ on the parabola,where $y = \sqrt{4ax}$. Since the rectangle $PQRS$ is inscribed in $T$ with $P, Q$ on the line $x = a$,the width of the rectangle is $(a - x)$ and the height is $2y$.Area of rectangle $A = 2y(a - x) = 2(2\sqrt{ax})(a - x) = 4\sqrt{a} \cdot x^{1/2}(a - x) = 4\sqrt{a}(ax^{1/2} - x^{3/2})$.To maximize the area,differentiate with respect to $x$:$\frac{dA}{dx} = 4\sqrt{a} \left( \frac{a}{2\sqrt{x}} - \frac{3}{2}\sqrt{x} ight) = 0 \implies \frac{a}{2\sqrt{x}} = \frac{3\sqrt{x}}{2} \implies a = 3x \implies x = \frac{a}{3}$.Then $y^2 = 4a(a/3) = 4a^2/3$,so $y = \frac{2a}{\sqrt{3}}$.The maximum area of the rectangle is $A = 2y(a - x) = 2 \left( \frac{2a}{\sqrt{3}} ight) \left( a - \frac{a}{3} ight) = \frac{4a}{\sqrt{3}} \cdot \frac{2a}{3} = \frac{8a^2}{3\sqrt{3}}$.The ratio is $\frac{	ext{area}(PQRS)}{	ext{area}(T)} = \frac{8a^2 / 3\sqrt{3}}{8a^2 / 3} = \frac{1}{\sqrt{3}}$.

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