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A normal drawn to the parabola y^2=4a x ...

A normal drawn to the parabola `y^2=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

A

A. `(8a,4sqrt(2)a)`

B

B. `(8a,4a)`

C

C. `(2a,-2sqrt(2)a)`

D

D. `(2a,2sqrt(2)a)`

Text Solution

Verified by Experts

The correct Answer is:
C, D
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)`.
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