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Let $\mathrm{AB}$ be the latus rectum of the parabola $y^{2}=4 a x$ in the $x y$-plane. Let $T$ be the region bounded by the finite arc $\mathrm{AB}$ of the parabola and the line segment $\mathrm{AB}$. A rectangle PQRS of maximum possible area is inscribed in $\mathrm{T}$ with $\mathrm{P}, \mathrm{Q}$ on line $\mathrm{AB}$, and $\mathrm{R}, \mathrm{S}$ on arc $\mathrm{AB}$. Then area(PQRS)/area(T) equals
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Verified Answer
The correct answer is:
$\frac{1}{\sqrt{3}}$
$\begin{array}{l}
T=\int_{0}^{a} 2 \sqrt{a} \sqrt{x} \cdot d x=\frac{8 a^{2}}{3} \\
t_{1}=t_{2}
\end{array}$
Area of PQRS
$=2 \mathrm{a}\left|\mathrm{t}_{1}-\mathrm{t}_{2}\right| \times\left|\mathrm{a}-\mathrm{at}_{1}^{2}\right|$
But $\mathrm{t}_{1}=-\mathrm{t}_{2}$
$=4 \mathrm{a}^{2} \mathrm{t}_{1}\left(1-\mathrm{t}_{1}^{2}\right)$
Differentiation with respect $\mathrm{t}_{1}$
We will get $\mathrm{t}_{1}^{2}=\frac{1}{3}$
Now put $\mathrm{t}_{1}=\frac{1}{\sqrt{3}}$ get Area of $\mathrm{PQRS}=\frac{8 \mathrm{a}^{2}}{3 \sqrt{3}}$
ratio Becomes $\frac{1}{\sqrt{3}}$
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