A normal drawn to the parabola `=4a x` meets the curve again at `Q` such that the angle subtended by `P Q` at the vertex is `90^0dot` Then the coordinates of `P` can be `(8a ,4sqrt(2)a)` (b) `(8a ,4a)` `(2a ,-2sqrt(2)a)` (d) `(2a ,2sqrt(2)a)`

Options

A. `(8a,4sqrt(2)a)`
B. (8a,4a)
C. `(2a,-2sqrt(2)a)`
D. `(2a,2sqrt(2)a)`

Answer

C, D

Explanation

3,4
`t_(2)=-t_(1)-(2)/(t_(1))`
Also, `(2at_(1))/(at_(1)^(2))xx(2at_(2))/(at_(2)^(2))=-1`
`ort_(1)t_(2)=-4`
`:.(-4)/(t_(1))=-t_(1)-(2)/(t_(1))`
`ort_(1)^(2)+2=4andt_(1)=pmsqrt(2)`
So, the point can be `(2a,pm2sqrt(2)a)`.