Question

Locus of a point through which three normals of parabola $y^2=4ax$ are passing, two of which are making angles $\alpha$ and $\beta$ with positive $x-$ axis and $\tan \alpha \cdot \tan \beta =2$ is

• A. $y(y^2-2ax)=0$
• B. $y(y^2+2ax)=0$
• C. $y(y^2-4ax)=0$
• D. $y(y^2-ax)=0$

Answer 1

The correct option is C $y(y^2-4ax)=0$

Let $(h,k)$ be a point on required locus.
Equation of normal in slope form is
$y=mx-2am-a{m}^3$
$\because$ Normal passes through $(h,k)$,
$\Rightarrow a{m}^3+(2a-h)m+k=0\cdots \left(i\right)$
Let slopes of the normal be $m_1,m_2,m_3$, then
$m_1{m}_2{m}_3=-\frac{k}{a}$
Given $m_1{m}_2=\tan \alpha \tan \beta =2$
$\Rightarrow m_3=-\frac{k}{2a}$
As, $m_3$ satisfies equation $\left(i\right)$
$-a\left(\frac{k^3}{8a^3}\right)-(2a-h)\frac{k}{2a}+k=0$

So, the required equation of the locus is
$y(y^2+4a(2a-x)-8a^2)=0\therefore y(y^2-4ax)=0$