Question

The equation of the locus of the point of intersection of two normals to the parabola $y^2=4ax$ which are perpendicular to each other is

• A. $y^2=a(x-3a)$
• B. $y^2=a(x+3a)$
• C. $y^2=a(x+2a)$
• D. $y^2=a(x-2a)$

Answer 1

The correct option is A $y^2=a(x-3a)$

Let $P(x_1,y_1)$ be the point of intersectin of the two perpendicular normals at $A\left(t_1\right),B\left(t_2\right)$ on the parabola $y^2=4ax$
let $t_3$ be the foot of the third normal through P.Equation of a normal at t to the parabola is $y+xt=2at+a{t}^3$
If this normal passes through P then $y_1+x_{1}t=2at+a{t}^3\Rightarrow a{t}^3+(2a-x_1)t-y_1=0\to \left(1\right)$
Now $t_1,t_2,t_3$ are the roots of (1). $\therefore t_1{t}_2{t}_3=\frac{y_1}{a}$
Slope of the normal at $t_1$ is $-t_1$ , slope of normal at $t_2$ is $-t_2$
Normals at $t_1$ and $t_2$ are perpendicular $\Rightarrow (-t_1)(-t_2)=-1\Rightarrow t_1{t}_2{t}_3=-t_3\Rightarrow \frac{y_1}{a}=-t_3\Rightarrow t_3=\frac{y_1}{a}$
$t_3$ is a root of (1) $\Rightarrow a{(-\frac{y_1}{a})}^3+(2a-x_1)(-\frac{y_1}{a})-y_1=0\Rightarrow -\frac{y_1^3}{a^2}-\frac{(2a-x_1)y_1}{a}-y_1=0$
$\Rightarrow y_1^2+a(2a-x_1)+a^2=0\Rightarrow y_1^2=a(x_1-3a)\therefore$ The locus of P is $y^2=a(x-3a)$