If X is the foot of the directrix on the a parabola. PP' is a double ordinate of the curve and PX meets the curve again in Q. Then prove that PQ passes through the focus of the parabola.

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.