Find the locus of the point of intersection of two normals to a parabolas which are at right angles to one another.

Explanation

The equation of the normal to the parabola `y^(2)=4ax` is
`y=mx-2am-am^(3)`
It passes through the point (h,k) if
`k=mh-2am-am^(3)`
`or" "am^(2)+m(2a-h)+k=0` (1)
Let the roots of the above equation be `m_(1),m_(2),andm_(3)`.
Let normals having slopes `m_(1)andm_(2)` be perpendicular.
so, `m_(1)m_(2)=-1`
From (1), `m_(1)m_(2)m_(3)=-k//a`.
Since `m_(1)m_(2)-1,m_(3)=k//a`.
Since `m_(3)` is a root of (1), we have
`a((k)/(a))^(3)+(k)/(a)(2a-h)+k=0`
`or" "k^(2)+a(2a-h)+a^(2)=0`
`or" "k^(2)=a(h-3a)`
Hence, the locus of (h,k) is
`y^(2)=a(x-3a)`