Question

Three normals with slopes $m_1,m_2$ and $m_3$ are drawn from a point $P$ not on the axis of the parabola $y^2=4x.$ If $m_1{m}_2=\alpha$, results in the locus of $P$ being a part of the parabola, then the value of $\alpha$ is

• A. $1$
• B. $2$
• C. $\frac{1}{2}$
• D. None of these

Answer 1

Answer: (B)

$2$

Equation of normal $y=mx-2am-{am}^2,$ passes through $P(h,k)m^3+(2-h)m+k=0;$

Let the roots be $m_1,m_2,m_3$, then

$m_1.m_2.m_3=-k$ $\Rightarrow \alpha .m_3=-k$

$\therefore m_3=\frac{-k}{\alpha }$

$\therefore {\left(\frac{-k}{\alpha }\right)}^3+(2-h)\left(\frac{-k}{\alpha }\right)+k=0$

$k^3+(2-h)k{\alpha }^2-k{\alpha }^3=0$

$\therefore$ Locus of $P$ is $y^2+(2-x){\alpha }^2-{\alpha }^3=0$

i.e., $y^2={\alpha }^{2}x-2{\alpha }^2+{\alpha }^3;$

If it lies on the parabola $y^2=4x$

Then ${\alpha }^2=4$ and $-2{\alpha }^2+{\alpha }^3=0$

${\alpha }^2(\alpha -2)=0$

$\therefore \alpha =2$