Question

The three normals from a point to the parabola $y^2=4ax$ cut the axes in points, whose distances from the vertex are in A.P., then the locus of the point is

• A. $27a{y}^2=2(x-2a{)}^3$
• B. $27a{y}^3=2(x-2a{)}^2$
• C. $9a{y}^2=2(x-2a{)}^3$
• D. $9a{y}^3=2(x-2a{)}^2$

Answer 1

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

Let $P(x_1,y_1)$ be any point.
Equation of any normal is $y=mx-2am-a{m}^3$
If it passes through P then $y_1=m{x}_1-2am-a{m}^3$
$\Rightarrow a{m}^3+(2a-x_1)m_1+y_1=0$, which is cubic in m.
Let $m_1,m_2{m}_3$ be its roots. Then $m_1+m_2+m_3=0,m_1{m}_2+m_2{m}_3+m_3{m}_1=\frac{2a-x_1}{a}$
Normal meets the axis (y=0), where $0=mx-2am-a{m}^3\Rightarrow x=2a+a{m}^2$
$\therefore$ Distances of points from the vertex are $2a+a{m}_1^2,2a+a{m}_2^2,2a+a{m}_3^2$
If these are in A.P., then $2(2a+a{m}_2^2)=(2a+a{m}_1^2)+(2a+a{m}_3^2)\Rightarrow 2m_2^2=m_1^2+m_3^2$
$\Rightarrow 3m_2^2=m_1^2+m_2^2=(m_1+m_2+m_3{)}^2-2(m_1{m}_2+m_2{m}_3+m_3{m}_1)=\frac{-2(2a-x_1)}{a}$
$\therefore m_2^2=\frac{-2(2a-x_1)}{3a}$
But $y_1=m_2(x_1-2a-a{m}_2^2)\Rightarrow y_1^2=m_2^2(x_1-2a-a{m}_2^2{)}^2=\frac{(x_1-2a{)}^3}{27a}$
Locus of P is $27a{y}^2=2(x-2a{)}^3$.