`PC` is the normal at P to the parabola `y^2=4ax, C` being on the axis. `CP` is produced outwards to Q so that `PQ =CP` ; show that the locus of Q is a parabola.

Options

A. `(a,0)`
B. `(-a,0)`
C. `(-2a,0)`
D. `(2a,0)`

Answer

D

Explanation

For parabola `y^(2) = 4ax`, normal at `P(at^(2),2at)` is `y =- tx +2at +at^(3)`

Putting `y =0`, we get
`tx = 2at + at^(3)`
`rArr x = 2a + at^(2) =a (2+t^(2))`
`:. C -= (2a + at^(2),0)`
Since, `PQ = CP,P` is the mid point of QC.
`:. h + 2a +at^(2) = 2at^(2)` (i)
and `k = 4at`
`rArr t = k//4a`
Putting value of t in (i), we get
`h + 2a =a ((k)/(4a))^(2)`
`rArr h +2a = (k^(2))/(16a)`
`rArr y^(2) = 16a (x+2a)`
which has focus at `(2a,0)`.