Let `S` is the focus of the parabola `y^2 = 4ax` and `X` the foot of the directrix, `PP'` is a double ordinate of the curve and `PX` meets the curve again in `Q.` Prove that `P'Q` passes through focus.

Explanation

Consider parabola `y^(2)=4ax`
`X-=(-a,0)`
Let point P be `(at^(2),2at)`.
`:." "P^(')-=(at^(2),-2at)`
Equation of line PX is
`y=(2at-0)/(at^(2)+a)(x+a)`
`or(1+t^(2))y=2t(x+a)` (1)
`(4t^(2)(x+a)^(2))/((1+t^(2))^(2))=4ax`
`rArr" "t^(2)(x+a)^(2)=ax(1+t^(2))^(2)`
`rArr" "t^(2)(x^(2)+a^(2)+2ax)=a(t^(4)+1+2t^(2))x`
`rArr" "t^(2)x^(2)+t^(2)a^(2)=xat^(4)+ax`
`rArr" "t^(2)x^(2)-(a+at^(4))x+at^(2)=0`
`rArr" "xt(x-at^(2))-a(x-at^(2))=0`
`rArr" "(x-at^(2))(xt^(2)-a)=0`
`rArr" "x=at^(2),a//t^(2)`
Putting `x=a//t^(2)` into (1), we get
`(1+t^(2))y=2at((1)/(t^(2)+a))`
`:." "y=2a//t`.
Clearly, points Q and P' are extremities of the focal chord.
So, P'Q passes through the focus.