Question

Let $A\left(4,-4\right)$ and $B\left(9,6\right)$ be points on the parabola, $y^2=4x$. Let C be chosen on the arc $AOB$ of the parabola, where $O$ is the origin, such that the area of $\Delta ACB$ is maximum. Then, the area (in sq. units) of $\Delta ACB$, is

Answer 1

Answer: 31.25

Let the coordinates of $C$ is $\left(t^2,2t\right)$. Since, area of $\Delta ACB$
$=\frac{1}{2}\Vert \begin{array}{ccc}{t}^2& 2t& 1\\ 9& 6& 1\\ 4& -4& 1\end{array}\Vert$
$=\frac{1}{2}\left|t^2\left(6+4\right)-2t\left(9-4\right)+1\left(-36-24\right)\right|$
$=\frac{1}{2}\left|10t^2-10t-60\right|$
$=5\left|t^2-t-6\right|$
$=5\left|{\left(t-\frac{1}{2}\right)}^2-\frac{25}{4}\right|$ [Here, $t\infty \left(0,3\right)$]
For maximum area, $t=\frac{1}{2}$
Hence, maximum area $=\frac{125}{4}=31.25$ sq. units