Through the vertex O of a parabola `y^(2) = 4x` chords OP and OQ are drawn at right angles to one another. Show that for all position of P, PQ cuts the axis of the parabola at a fixed point.

Explanation

The given parabola is
`y^(2)=4x` …(i)
Let
`P=equiv(t_(1)^(2),2t_(1)),`
`Qequiv(t_(2)^(2),2t_(2)).`
Slope of
`OP=(2t_(1))/(t_(1)^(2))`
`=2/(t_(1))`
and slope of
`OQ=2/(t_(2))`
Since
`OPbotOQ,`
`4/(t_(1)t_(2))=-1`
or `t_(1)t_(2)=-4` ....(ii)
The equation of PQ is
`y(t_(1)+t_(2))=2(x+t_(1)t_(2))`
`!y(t_(1)-4/(t_(1)))=2(x-4)`[from (ii)]
`!2(x-4)-y(t_(1)-4/(t_(1)))=0`
`!L_(1)+lamdaL_(2)=0`
`therefore` variable line PQ passes through a fixed point which is point of intersection of
`L_(1)=0`
`&L_(2)=0`
i.e., (4,0)