Question

Let P and Q be distinct points on the parabola $y^2=2x$ such that a circle with PQ as diameter passes through the vertex O of the parabola. If P lies in the first quadrant and the area of the triangle $\Delta OPQ$ is $3\sqrt{2},$ then which of the following is (are) the coordinates of P ?

• A. $\left(4,2\sqrt{2}\right)$
• B. $\left(9,3\sqrt{2}\right)$
• C. $\left(\frac{1}{4},\frac{1}{\sqrt{2}}\right)$
• D. $\left(1,\sqrt{2}\right)$

Answer 1

Answer: (A), (D)

Let coordinates of P and Q are P$\left(2t_1^2,2t_1\right),Q\left(2t_2^2,2t_2\right)$
As the circle with PQ as diameter passes through the vertex O.
So $\angle POQ=90^{\circ }$
$m_{OP}\times m_{OQ}=-1$
$\frac{2t_1}{2t_1^2}\times \frac{2t_2}{2t_2^2}=-1\Rightarrow t_1{t}_2=-1$
Now Area of $\Delta OPQ$ will be
$\frac{1}{2}\left|\begin{array}{ccc}1& 0& 0\\ 1& 2t_1^2& 2t_1\\ 1& 2t_2^2& 2t_2\end{array}\right|=3\sqrt{2}$
$\Rightarrow 4|t_1{t}_2\left(t_1-t_2\right)|=6\sqrt{2}<br/>\Rightarrow \left|t_1-t_2\right|=\frac{3}{\sqrt{2}}$
$\Rightarrow t_1+\frac{1}{t_1}=\frac{3}{\sqrt{2}}$
$\Rightarrow \sqrt{2}{t}_1^2-3t_1+\sqrt{2}=0$
$\Rightarrow \left(\sqrt{2}{t}_1-1\right)\left(t_1-\sqrt{2}\right)=0$
$\Rightarrow t_1=\frac{1}{\sqrt{2}},\sqrt{2}$
$\therefore$ Coordinates of P can be $\left(4,2\sqrt{2}\right),\left(1,\sqrt{2}\right)$