Find the locus of the middle points of the chords of the parabola `y^2=4a x` which subtend a right angle at the vertex of the parabola.

Explanation

The equation of the chord of the parabola whose middle point is `(alpha,beta)` is
`ybeta-2a(x+alpha)=beta^(2)-4aalpha`
`!ybeta-2ax=beta^(2)-2aalpha`
or `(ybeta-2ax)/(beta^(2)-2aalpha)=1` …(i)
Now, the equation of the pair of the lines OP and OQ joining the origin O i.e. the vertex to the points of intersection P and Q of the chord with the parabola `y^(2) = 4ax` is obtained by making the equation homogeneous by means of (i). Thus the equation of lines OP and OQ is
`y^(2)=(4ax(ybeta-2ax))/(beta^(2)-2aalpha)`
`!y^(2)(beta^(2)-2aalpha)-4abetaxy+8a^(2)x^(2)=0`
If the lines OP and OQ are at right angles, then the coefficient of `x^(2)+` the coefficient of `y^(2)=0`. Therefore, `beta^(2)-2aalpha+8a^(2)=0!beta^(2)=2a(alpha-4a)`. Hence the locus of `(alpha,beta)` is `y^(2)=2a(x-4a)`