Let PQ be a chord of the parabola $y^{2} = 4 x$ . A circle drawn with PQ as diameter passes through the vertex ‘V’ of the parabola. If the area of triangle PVQ is 20 then coordinates of P are

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

$(- 16, 8)$

$(- 16, - 8)$

$(8, 16)$

$(16, - 8)$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} S l o p e o f \bar{P V} X s l o p e o f \bar{Q V} = - 1; \\ P (a t_{1}, 2 a t_{1}), Q (a {t_{2}}^{2}, 2 a t_{2}) \Rightarrow t_{1} t_{2} = - 4 \\ A r e a : \frac{1}{2} | x_{1} y_{2} - x_{2} y_{1} | = 20 \Rightarrow | t_{1} t_{2} (t_{2} - t_{1}) | = 20 \\ \Rightarrow | - 4 (\frac{- 4}{t_{1}} - t_{1}) | = 20 \\ \Rightarrow \frac{4}{t_{1}} + t_{1} = 5; \frac{4}{t_{1}} + t_{1} = - 5 \\ \Rightarrow {t_{1}}^{2} - 5 t_{1} + 4 = 0 \Rightarrow t_{1} = 1; t_{1} = 4 t h e n t h e r e i s n o a n s w e r i n t h e g i v e n o p t i o n . \\ \Rightarrow {t_{1}}^{2} + 5 t_{1} + 4 = 0 \Rightarrow t_{1} = - 1; t_{1} = - 4 \\ I f w e t a k e t_{1} = - 4 ∴ P (16, - 8) . \end{array}$