Let P and Q be distinct points on the parabola $y^{2} = 2 x$ 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 $Δ O P Q$ is $3 \sqrt{2}$ , then which of the following is (are) the coordinates of P?

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$(4, 2 \sqrt{2})$

$(\frac{1}{4}, \frac{1}{\sqrt{2}})$

$(1, \sqrt{2})$

$(9, 3 \sqrt{2})$

answer is A, D.

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Detailed Solution

$\begin{array}{l} L e t \begin{array}{l} P (a t_{1}^{2}, 2 a t_{1}) & Q (a t_{2}^{2}, 2 a t_{2}) b e a n y t w o p o i n t s o n t h e p a r a b o l a y^{2} = 2 x \end{array} \\ \Rightarrow 4 a = 2 a = \frac{1}{2} \\ N o w t h e d i a m e t r i c f o r m o f e q u a t i o n o f t h e c i r c l e i s g i v e n b y \\ (x - a t_{1}^{2}) (x - a t_{2}^{2}) + (y - 2 a t_{1}) (y - 2 a t_{2}) = 0 \\ Since the circle passes through (0, 0), \\ \Rightarrow a^{2} t_{1}^{2} t_{2}^{2} + 4 a^{2} t_{1} t_{2} = 0 \\ \Rightarrow a^{2} t_{1} t_{2} (t_{1} t_{2} + 4) = 0 \\ \Rightarrow t_{2} = \frac{- 4}{t_{1}} \to (1) \end{array} \overset{\leftrightarrow}{}$

$\begin{array}{l} g i v e n a r e a o f Δ O P Q = 3 \sqrt{2} \\ \Rightarrow \frac{1}{2} |a t_{1}^{2} \times 2 a t_{2} - 2 a t_{1} a t_{2}^{2}| = 3 \sqrt{2} \\ \Rightarrow |\frac{2 a^{2} t_{1} t_{2}}{2}| |t_{1} - t_{2}| = 3 \sqrt{2} \\ \Rightarrow |t_{1} - t_{2}| = 3 \sqrt{2} \end{array}$

$\begin{array}{l} |t_{1} - \frac{4}{t_{1}}| = 3 \sqrt{2} (b y (1)) \\ \Rightarrow t_{1} - \frac{4}{t_{1}} = 3 \sqrt{2}, t_{1} - \frac{4}{t_{1}} = - 3 \sqrt{2} \\ \Rightarrow t_{1}^{2} - 3 \sqrt{2} t_{1} - 4 = 0, t_{1}^{2} + 3 \sqrt{2} t_{1} - 4 = 0 \\ \Rightarrow t_{1}^{2} - 2 \sqrt{2} t_{1} - \sqrt{2} t_{1} - 4 = 0 \\ \Rightarrow t_{1} (t_{1} - 2 \sqrt{2}) - \sqrt{2} (t_{1} - 2 \sqrt{2}) = 0 \\ \Rightarrow t_{1} = \sqrt{2} (or) 2 \sqrt{2} \\ \Rightarrow P (a t_{1}^{2}, 2 a t_{1}) = (1, \sqrt{2}) (or) (4, 2 \sqrt{2}) \end{array}$