JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Parabola

The area of triangle $PXQ$ is largest when $X$ is the point of the parabola whose tangent is parallel to the chord $PQ$. The chord through $P(4,-4)$ and $Q(9,6)$ has slope $\frac{6-(-4)}{9-4}=2$. For $y^2=4x$, $\frac{dy}{dx}=\frac{2}{y}$, so the tangent has slope $2$ where $y=1$, giving $x=\frac{1}{4}$ and $X=\left(\frac{1}{4},1\right)$. The area with vertices $P(4,-4)$, $X\left(\frac{1}{4},1\right)$ and $Q(9,6)$ is $\frac{1}{2}\left|4(1-6)+\frac{1}{4}(6+4)+9(-4-1)\right|=\frac{1}{2}\left|-20+\frac{10}{4}-45\right|=\frac{125}{4}$ square units.