Question

If from a point $P$, $3$ normals are drawn to parabola $y^2=4ax$, then the locus of $P$ such that one of the normal is angular bisector of other two normals is

• A. $(2x-a)(x-5a{)}^2=27a{y}^2$
• B. $(2x-a)(x+5a{)}^2=27a{y}^2$
• C. $(2x-a)(x-5a)=27a{y}^2$
• D. $(2x-a)(x+5a)=27a{y}^2$

Answer 1

The correct option is A $(2x-a)(x-5a{)}^2=27a{y}^2$

Equation of parabola is $y^2=4ax$
Equation of normal is
$y=mx-2am-a{m}^3$

Let the normal is drawn from $P(h,k)$
$\therefore k=mh-2am-a{m}^3\Rightarrow a{m}^3+m(2a-h)+k=0\cdots \left(1\right)$
For three distinct normals, the equation has three real and distinct roots as $m_1,m_2,m_3$
Sum of roots is
$m_1+m_2+m_3=0\cdots \left(2\right)$

If $\alpha ,\beta ,\gamma$ are the angles made by the normals with the $x-$axis, then
$\tan \alpha =m_1,\tan \beta =m_2,\tan \gamma =m_3$
$\because$ One normal is angle bisector of other two normals
$\therefore \alpha -\beta =\beta -\gamma \Rightarrow 2\beta =\alpha +\gamma \Rightarrow \tan 2\beta =\tan (\alpha +\gamma )\Rightarrow \frac{2m_2}{1-m_2^2}=\frac{m_1+m_3}{1-m_1{m}_3}\Rightarrow \frac{2m_2}{1-m_2^2}=\frac{-m_2}{1-m_1{m}_3}\left[\text{using}\right(2\left)\right]\Rightarrow m_2^2+2m_1{m}_3-3=0\cdots \left(3\right)$

Also,
$m_1{m}_2+m_2{m}_3+m_3{m}_1=\frac{2a-h}{a}\Rightarrow -m_2^2+m_1{m}_3=\frac{2a-h}{a}\cdots \left(4\right)\left[\text{using}\right(2\left)\right]$

From equations $\left(3\right)$ and $\left(4\right)$, we get
$m_2^2=\frac{2h-a}{3a},m_1{m}_3=\frac{5a-h}{3a}$

Also,
$m_1{m}_2{m}_3=-\frac{k}{a}\therefore \left(\frac{2h-a}{3a}\right){\left(\frac{5a-h}{3a}\right)}^2=\frac{k^2}{a^2}$

Hence, the locus is
$(2x-a)(x-5a{)}^2=27a{y}^2$