Question

Let $P(4,-4)$ and $Q(9,6)$ be two points on the parabola $y^2=4x$ and let $X$ be any point on the arc $POQ$ of this parabola, where $O$ is the vertex of this parabola, such that the area of $△PXQ$ is maximum. Then $4$ times this maximum area (in sq. units) is

Answer 1

Cosider the parabola $y^2=4x$
Let the coordinates of point $X$ is $(t^2,2t)$


Area of $△PXQ$
$=\frac{1}{2}\left[t^2\right(-4-6)-2t(4-9)+1(24+36\left)\right]$
$=\frac{1}{2}[-10t^2+10t+60]$
$=5(-t^2+t+6)$
$=5[\frac{25}{4}-{(t-\frac{1}{2})}^2]$
Hence area will be maximum when $t=\frac{1}{2}$
$\therefore$ maximum area $=5\left(\frac{25}{4}\right)$
$=\frac{125}{4}\text{sq. units}$