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 $\Delta PXQ$ is maximum. Then this maximum area (in sq. units) is:

• A. $\frac{125}{2}$
• B. $\frac{625}{4}$
• C. $\frac{125}{4}$
• D. $\frac{75}{2}$

Answer 1

The correct option is C $\frac{125}{4}$

Cosider the parabola $y^2=4x$

Let the coordinates of point $X$ is $(t^2,2t)$
$\therefore$ area of $△PXQ=\left|\begin{array}{ccc}{t}^2& 2t& 1\\ 4& -4& 1\\ 9& 6& 1\end{array}\right|$
$\Rightarrow A=\frac{1}{2}\left[t^2\right(-4-6)-2t(4-9)+1(24+36\left)\right]$
$=\frac{1}{2}[-10t^2+10t+60]$
$\Rightarrow A=5(-t^2+t+6)$
$\Rightarrow \frac{dA}{dt}=5(-2t+1)=0$
$\Rightarrow t=\frac{1}{2}$
$\therefore$ minimum area $=5(-\frac{1}{4}+\frac{1}{2}+6)$
$=\frac{125}{4}\text{sq. units}$