The equation of the locus of the point of intersection of two normals to the parabola y2=4ax which are perpendicular to each other is
y2=a(x−3a)
Let P(x1,y1) be the point of intersection of the two perpendicular normals at A(t1), B(t2) on the parabola
y2=4ax.
Let t3 be the foot of the third normal through P
Equation of a normal at t to the parabola is y+xt=2at+at3
If this normal passes through P then
y1+x1t=2at+at3⇒at3+(2a−x1)t−y1=0 ...... (1)
Now t1, t2, t3 are the roots of (1)
∴t1 t2 t3=y1a
Slope of the normal at t1 is –t1
Slope of the normal at t2 is –t2
Normals at t1 and t2 are perpendicular ⇒(−t1)(−t2)=−1⇒t1 t2=−1⇒t1 t2 t3=−t3
⇒y1a=−t3⇒t3=−y1a
t3 is a root of (1) ⇒a(−y1a)3+(2a−x1)(−y1a)−y1=0⇒−y31a2−(2a−x1)y1a−y1=0
⇒y21+a(2a−x1)+a2=0⇒y21=a(x1−3a)
∴ The locus of P is y2=a(x−3a)