Consider the parabola y^{2}=4x. Let P and Q b | vedclass

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(C) The area of $\Delta PQR$ is maximized when the distance from $R$ to the line $PQ$ is maximum. Let $R$ be $(t^{2}, 2t)$. The line $PQ$ passes throu

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$R\left(\frac{1}{4}, 1\right)$

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(C) The area of $\Delta PQR$ is maximized when the distance from $R$ to the line $PQ$ is maximum. Let $R$ be $(t^{2}, 2t)$. The line $PQ$ passes through $P(4, -4)$ and $Q(9, 6)$. The slope of $PQ$ is $m = \frac{6 - (-4)}{9 - 4} = \frac{10}{5} = 2$. The equation of line $PQ$ is $y - 6 = 2(x - 9) \Rightarrow 2x - y - 12 = 0$. The perpendicular distance from $R(t^{2}, 2t)$ to $2x - y - 12 = 0$ is $d = \frac{|2t^{2} - 2t - 12|}{\sqrt{2^{2} + (-1)^{2}}} = \frac{|2(t^{2} - t - 6)|}{\sqrt{5}} = \frac{2|t - 3||t + 2|}{\sqrt{5}}$. For $R$ to be on the arc between $P$ and $Q$, the parameter $t$ must lie between $-2$ and $3$. Let $f(t) = t^{2} - t - 6$. To maximize the distance, we find the critical point of $f(t)$ by setting $f'(t) = 2t - 1 = 0$, which gives $t = \frac{1}{2}$. At $t = \frac{1}{2}$, the coordinates of $R$ are $\left(\left(\frac{1}{2}\right)^{2}, 2\left(\frac{1}{2}\right)\right) = \left(\frac{1}{4}, 1\right)$.

Consider the parabola $y^{2}=4x$. Let $P$ and $Q$ be points on the parabola where $P(4, -4)$ and $Q(9, 6)$. Let $R$ be a point on the arc of the parabola between $P$ and $Q$. Then, the area of $\Delta PQR$ is largest when

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