Normals at P, Q, R are drawn to `y^(2)=4x` which intersect at (3, 0). Then, area of `DeltaPQR`, is

Options

A. `2//5`
B. `1//2`
C. `5//2`
D. 2

Answer

D

Explanation

The equation of a normal to the parabola `y^(2)=4x`, is `y=mx-2m-m^(3)`
If it passes through (3, 0), then
`0=3m-2m-m^(3)`
`rArr" "m(1-m)^(2)=0rArrm=0+-1`

The coordinates the feet of the normals drawn from (3, 0) are `P(m_(1)^(2),-2m_(1)), Q(m_(2)^(2), -2m_(2))" and "R(m_(3)^(2),-2m_(3))` where `m_(1),m_(1),m_(3)` are roots of equation (i).
Let `m_(1)=0, m_(2)=-1" and "m_(3)=1`. Then, coordinates of P, Q and R are (0, 0), (1, 2) and (1, -2) respectively.
`:." Area of "DeltaPQR=1/2xxQRxxPM,"where PM is perpendicular from P on QR"`
`rArr" Area of "DeltaPQR=1/2xx4xx1=2`