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 $ΔOPQ$ 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 $∠POQ = 90^{\circ}$
$m_{OP} × m_{OQ} = -1$
$\frac{2t_{1}}{2t^{2}_{1}}\times \frac{2t_{2}}{2t^{2}_{2}} = -1\,\, \Rightarrow t_{1}t_{2} = -1$
Now Area of $ΔOPQ$ will be
$\frac{1}{2}\begin{vmatrix}1&0&0\\ 1&2t^{2}_{1}&2t_{1}\\ 1&2t^{2}_{2}&2t_{2}\end{vmatrix} = 3\sqrt{2}$
$⇒ 4|t_{1} t_{2} \left(t_{1} - t_{2}\right)| = 6\sqrt{2}
\Rightarrow \left|t_{1}-t_{2}\right| = \frac{3}{\sqrt{2}}$
$\Rightarrow t_{1}+\frac{1}{t_{1}} = \frac{3}{\sqrt{2}}$
$\Rightarrow \sqrt{2} t^{2}_{1} - 3t_{1} +\sqrt{2} = 0$
$\Rightarrow \left(\sqrt{2}t_{1}-1\right)\left(t_{1}-\sqrt{2}\right) = 0$
$\Rightarrow t_{1} = \frac{1}{\sqrt{2}}, \sqrt{2}$
$∴$ Coordinates of P can be $\left(4, 2 \sqrt{2} \right), \left(1, \sqrt{2} \right)$