JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Parabola

Parametrise the parabola $y^2=4x$ as $(t^2,2t)$; then $A(4,-4)$ is $t=-2$ and $B(9,6)$ is $t=3$. The triangle $ACB$ has greatest area when the tangent at $C$ is parallel to chord $AB$. The slope of $AB$ is $\frac{6-(-4)}{9-4}=2$, and the tangent slope at parameter $t$ is $\frac{1}{t}$, so $\frac{1}{t}=2$ gives $t=\frac{1}{2}$ and $C=\left(\frac{1}{4},1\right)$. The area with vertices $A(4,-4)$, $B(9,6)$, $C\left(\frac{1}{4},1\right)$ is $\frac{1}{2}\left|4(6-1)+9(1+4)+\frac{1}{4}(-4-6)\right|=\frac{1}{2}\left|20+45-2.5\right|=31\frac{1}{4}$ square units.