# Geo5.1.42

Prove that the locus of the point of intersection of two perpendicular normals to the parabola $y^{2}=4ax\,$ is the parabola $y^{2}=a(x-3a)\,$

Let $P(x_{1},y_{1})\,$ be the ppoint of intersection of the normals at $A(t_{1}),B(t_{2})\,$

Equation of the normal at t is $y+tx=2at+at^{3}\,$

If it passes through (x1,y1)

$y_{1}+tx_{1}=2at+at^{3},at^{3}+(2a-x_{1})t-y_{1}=0\,$

If t1,t2,t3 are the roots of the above equation, $t_{1}t_{2}t_{3}={\frac {y_{1}}{a}}\,$.

Let this be equation 2

Slope of the normal at t1=-t1.

Slope of the normal at t2=-t2.

Given PA is perpendicular to PB $(-t_{1})(-t_{2})=-1,t_{1}t_{2}=1\,$.equation 3

From 2 and 3

$t_{3}={\frac {-y}{a}}\,$

Since the normal at t3 also passes thro' $P(x_{1},y_{1})\,$

$y_{1}+x_{1}t_{3}=2at_{3}+at_{3}^{{3}}\,$. equation 4.

$y_{1}+x_{1}({\frac {-y_{1}}{a}})=2a({\frac {-y_{1}}{a}}+a({\frac {-y_{1}^{{3}}}{a^{3}}}\,$

$1-{\frac {x_{1}}{a}}=-2-{\frac {y_{1}^{{2}}}{a^{2}}}\,$

$y_{1}^{{2}}=a(x_{1}-3a)\,$

Hence the locus of P is

$y^{2}=a(x-3a)\,$