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 intersection 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

$y_1+x_{1}t=2at+a{t}^3\Rightarrow a{t}^3+(2a-x_1)t-y_1=0$ ...... (1)

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 the normal at $t_2$ is $–t_2$

Normals at $t_1$ and $t_2$ are perpendicular $\Rightarrow (-t_1)(-t_2)=-1\Rightarrow t_1{t}_2=-1\Rightarrow t_1{t}_2{t}_3=-t_3$

$\Rightarrow \frac{y_1}{a}=-t_3\Rightarrow t_3=-\frac{y_1}{a}$

$t_3$ 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 $y^2=a(x-3a)$