Let PQ be a chord of the parabola `y^2=4x`. A circle drawn with PQ as a diameter passes through the vertex V of theparabola. If `ar(Delta PVQ)=20` sq unit then the coordinates of P are

Options

A. (16,8)
B. (16,-8)
C. (9,6)
D. (9,-6)

Answer

A, B

Explanation

1,2
Slope of PV `=(2t-0)/(t^(2)-0)=(2)/(t)`
Therefore, the equation of QV is
`y=-(t)/(2)x`

Solving it with `y^(2)-4x," we get "Q-=(16//t^(2),-8//t)`. Now
`ar(DeltaPVQ)=(1)/(2)PV*VQ=20` (Given)
`:.PV^(2)*VQ^(2)=40^(2)`
`or{(t^(2))^(2)+(2t)^(2)}{((16)/(t^(2)))^(2)+((-8)/(t))}=40^(2)`
`ort^(2)(4+t^(2))(1)/(t^(2))((256)/(t^(2))+64)=40^(2)`
`or(256xx4)/(t^(2))+256+256+64t^(2)=40^(2)`
`or(t^(2)-16)(t^(2)-1)=0`
`:.t=pm4,pm1`