Question

Three normals are drawn from $P$ to the parabola $y^2=4ax$.

If two normals are perpendicular then the equation to the locus of $P$ is?

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

Answer 1

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

Equation of normal to parabola in slope form is $y=mx-2am-a{m}^3$

Let $(h,k)$ be the coordinates of P.

$\therefore k=mh-2am-a{m}^3$

Let $m_1,m_2,m_3$ be the three roots of above equation

$⟹m_1{m}_2{m}_3=-\frac{k}{a}$ ...$\left[1\right]$ ...(product of roots)

Let $m_{1}and{m}_2$ be the slope of the two perpendicular normals

$⟹m_1{m}_2=-1$

Substituting this in equation[1] we get,

$\therefore m_3=\frac{k}{a}$

$\therefore k=\frac{k}{a}h-2a\frac{k}{a}-a{\frac{k}{a}}^3$

$\therefore k=\left(\frac{k}{a}\right)h-2a\left(\frac{k}{a}\right)-a(\frac{k}{a}{)}^3$

$\therefore a^2=ha-2a^2-k^2$

$\therefore k^2=a(h-3a)$

$\therefore y^2=a(x-3a)$