Find the locus of the point of intersection of two normals to a parabola which are at right angles to one another.
The equation of the normal to the parabola y2=4ax is y=mx−2am−am3.It passes through the point (h,k) is
k=mh−2am−am3
am3+m(2a−h)+k=0 .....(1)
Let the roots of the above equation be m1,m2,m3. Let the perpendicular normals correspond to the values of m1 and m2 so that m1m2=−1.
From equation 1, m1m2m3=−ka
Since, m1m2=−1,m3=ka
Since m3 is a root of equation 1, we have
a(ka)3+ka(2a−h)+k=0
k2+a(2a−h)+a2=0
k2=a(h−3a)
Hence, the locus of (h,k) is y2=a(x−3a).