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

Answer 1

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

$y^2=4x$
$2y{y}^{\prime }=4$
$y^{\prime }=\frac{1}{t}=2,t=\frac{1}{2}$
Area $=\frac{1}{2}\left|\begin{array}{ccc}\frac{1}{4}& 1& 1\\ 9& 6& 1\\ 4& -4& 1\end{array}\right|=\frac{125}{4}$