Question

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

• A. $\angle PQR={90}^{\circ }$
• B. $R(4,4)$
• C. $R(\frac{1}{4},1)$
• D. $R(1,\frac{1}{4})$

Answer 1

The correct option is C $R(\frac{1}{4},1)$


Equation of $PQ$ is
$\frac{y+4}{6+4}=\frac{x-4}{9-4}$
$\Rightarrow 2x–y=12$

Distance $RL=\left|\frac{2t^2-2t-12}{\sqrt{5}}\right|$
Assuming
$f=\frac{\sqrt{5}RL}{2}\Rightarrow f=\pm (t^2-t-12)=(t-3)(t+2)\Rightarrow f=\{\begin{array}{c}(t-3)(t+2)t\in (-\infty ,-2]\cup [3,\infty )\\ -(t-3)(t+2)t\in (-2,3)\end{array}$

Differentiate w.r.t. $t$,
$f^{\prime }=\pm (2t-1)=0\Rightarrow t=\frac{1}{2}$
Checking for maxima and minima at $t=\frac{1}{2}$
$f=-(t-3)(t+2)\Rightarrow f^{\prime }=-(2t-1)\Rightarrow f^{\prime \prime }=-2<0$
So at $t=\frac{1}{2}$ it will give maxima,

Hence $R=(\frac{1}{4},1)$