Question

A pair of tangents are drawn to the parabola $y^2=4ax$ which are equally inclined to a straight line $y=mx+c$, whose inclination to the axis is $\alpha$ then locus of their point of intersection is

• A. $y=(x-a)\tan 2\alpha$
• B. $y=(x+a)\tan 2\alpha$
• C. $y=(x-a)\tan 2\alpha$
• D. $y=(x+a)\tan \alpha$

Answer 1

The correct option is A $y=(x-a)\tan 2\alpha$

Let the angle between the line and the tangents is $\theta$ and ${\alpha }_1\&{\alpha }_2$ are the incilnation of the tangents to the axis

$\therefore \alpha -{\alpha }_1=\theta ={\alpha }_2-\alpha \Rightarrow 2\alpha ={\alpha }_1+{\alpha }_2\tan 2\alpha =\frac{m_1+m_2}{1-m_1{m}_2}$
Let $(h,k)$ be the point from which pair of tangents are drawn
$\therefore$ $(h,k)$ lies on the
$y=mx+\frac{a}{m}\Rightarrow m^{2}h-mk+a=0m_1+m_2=\frac{k}{h},m_1{m}_2=\frac{a}{h}\therefore \tan 2\alpha =\frac{\frac{k}{h}}{1-\frac{a}{h}}\Rightarrow y=(x-a)\tan 2\alpha$