Question

If from a point $P$, $3$ normals are drawn, to parabola $y^2=4ax$. Then the locus of $P$ such that three normals intersect the $x-$ axis at points whose distances from vertex are in $A.P.$ is

• A. $27a{y}^2=2(x+a{)}^3$
• B. $27a{y}^2=2(x-a{)}^3$
• C. $27a{y}^2=2(x+2a{)}^3$
• D. $27a{y}^2=2(x-2a{)}^3$

Answer 1

The correct option is D $27a{y}^2=2(x-2a{)}^3$

Equation of parabola $y^2=4ax$
Equation of normal is
$y=mx-2am-a{m}^3$
Let the normal drawn through $P(h,k)$
$\therefore k=mh-2am-a{m}^3$
For three distinct normals it has three real roots as
$m_1,m_2,m_3$

For normal to intersect with $x-$axis, $y=0$
$mx-2am-a{m}^3=0\Rightarrow x=2a+a{m}^2$
Hence values of $x$ for $m_1,m_2,m_3$ are in $A.P.$
$\therefore 2a+a{m}_1^2+2a+a{m}_3^2=2(2a+a{m}_2^2)\Rightarrow m_1^2+m_3^2=2m_2^2\cdots \left(1\right)$

But $m_1+m_2+m_3=0$ and
$m_1{m}_2+m_2{m}_3+m_1{m}_3=\frac{2a-h}{a}\Rightarrow m_1{m}_3-m_2^2=\frac{2a-h}{a}\cdots \left(2\right)$
And
$m_1+m_3=-m_2\Rightarrow m_1^2+m_3^2+2m_1{m}_3=m_2^2$
Using equation $\left(1\right)$, we get
$\Rightarrow 2m_1{m}_3+m_2^2=0\cdots \left(3\right)$

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

Also,
$m_1{m}_2{m}_3=-\frac{k}{a}\therefore -\frac{2(2a-h)}{3a}\cdot \frac{(2a-h{)}^2}{9a^2}=\frac{k^2}{a^2}$

Hence locus is
$27a{y}^2=-2(2a-x{)}^3\therefore 27a{y}^2=2(x-2a{)}^3$