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

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