Question

The locus of point of intersection of two normals drawn to the parabola $y^2=4ax$ which are at right angles is

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

Answer 1

Answer: (A)

The given parabola is $y^2=4ax$

Let $t$ denote the parameter, then the equation of the normal to this parabola is given by $y=-tx+2at+a{t}^3$

Now this normal passes through the point $(h,k)$ then

$k=-th+2at+a{t}^3\Rightarrow a{t}^3+t(2a-h)-k=0$ ....... $\left(A\right)$

Let $m_1,m_2,m_3$ be the roots of this equation.

Let the perpendicular normals correspond to the values of $m_1,m_2$

So, $m_1{m}_2=-1$

Now from equation $\left(A\right)$, $m_1{m}_2{m}_3=\frac{k}{a}$

$m_1{m}_2=-1$ gives $m_3=-\frac{k}{a}$

Since $m_3$ is a root of equation $\left(A\right)$, we have

$a{(-\frac{k}{a})}^3-\frac{k}{a}(2a-h)-k=0\Rightarrow k^2=a(h-3a)$

So locus of $(h,k)$ is $y^2=a(x-3a)$