Question

Find the locus of a point which is such that two of normals drawn from it to the parabola y^2=4ax are at right angles.

Answer 1

The locus of those points will not be on the parabola.
The question is to find the locus of point of intersection of two normals which are at right angles to each other.

Let P(h,k) be such a point.
Now, equation of the normal to the given parabola y²=4ax is,

y=mx−2am−am³

It passes through (h,k).So putting (h,k) in above equation we get,

k=mh−2am−am³

⇒am³ + (2a−h)m+k=0...........(i)
Let m₁,m₂and m₃ be its roots,which are also the slopes of the three normals from P. Since two normals are ⊥, then
m₁×m₂=−1
Also,m₁×m₂×m₃=−k/a
⇒m₃=k/a
Now, m₃ is arrot of equation (i),hence putting value of m₃ in equation (1),
a(k³/a³) + (2a−h)(k/a) + k=0

⇒k²+a(2a−h)+a²=0

⇒k²=a(h−3a)
hence locus of P(h,k) is
y²=a(x−3a)