Question

Let $P(4,-4)$ and $Q(9,6)$ be two points on the parabola $y^2=4x$. If $X$ is any point on the arc $POQ$ of the parabola, where $O$ is the vertex such that the area of $△PXQ$ is maximum, then $4$ times the 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 $△PXQ$
$=\frac{1}{2}\left|\begin{array}{cccc}4& t^2& 9& 4\\ -4& 2t& 6& -4\end{array}\right|=\frac{1}{2}|8t+4t^2+6t^2-18t-36-24|=|5t^2-5t-30|=5|t^2-t-6|=5|(t-3)(t+2)|$
Since, $X$ lies on the curve $POQ.$
Therefore, $t\in (-2,3)$, so
$=-5(t-3)(t+2)=5[\frac{25}{4}-{(t-\frac{1}{2})}^2]$
The maximum value of the area occurs when $t=\frac{1}{2}$
$A_{max}=\frac{125}{4}\Rightarrow 4A_{max}=125$