If X is the foot of the directrix on axis of the parabola. PP' is a double ordinate of the curve and PX meets the curve again in Q. Then prove that P' Q passes through fixed point which is

Options

A. vertex
B. focus
C. midpoint of vertex and focus
D. none of these

Answer

B

Explanation

(2) Consider parabola `y^(2)=4ax`
So, `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)
Solving this line with given parabola, we get
`(4t^(2)(x+a)^(2))/((l+t^(2))^(2))=4ax`
`rArrt^(2)(x+a)^(2)=ax(1+t^(2))^(2)`
`rArrt^(2)(x^(2)+a^(2)+2ax)=a(t^(4)+1+2t^(2))x`
`rArrt^(2)x^(2)+t^(2)a^(2)=xat^(4)+ax`
`rArrt^(2)x^(2)-(a+at^(4))x+a^(2)t^(2)=0`
`rArrxt^(2)(x-at^(2))-a(x-at^(2))=0`
`rArr(a-at^(2))(xt^(2)-a)=0`
`rArrx=at^(2),a//t^(2)`
Putting `x=a//t^(2)` in (1), we get
`(1+t^(2))y=2t((a)/(t^(2))+a)`
`:.y=2a//t`
Thus, point Q is `(a//t^(2),2a//t)`.
Clearly, points Q and P' are extremities of focal chord.
So, P' Q passes through the focus.