Question

A pair of tangents are drawn which are equally inclined to a straight line whose inclination to the axis is $\alpha$; prove that the locus of their point of intersection is the straight line
$y=(x-a)\tan 2\alpha$

Answer 1

Let the point of intersection be $P(h,k)$

Equation of tangent to $y^2=4ax$ in slope form is

$y=mx+\frac{a}{m}my=m^{2}x+a{m}^{2}x-my+a=0$

It passes through $P(h,k)$

$m^{2}h-mk+a=0$

The equation is quadractic in $m$

$\therefore m_1+m_2=-\frac{-k}{h}=\frac{k}{h}.....\left(i\right)m_1{m}_2=\frac{a}{h}.........\left(ii\right)$

Slope of line which makes equal angle with the tangents $=\tan \alpha$

$\frac{m_1-\tan \alpha }{1+m_1\tan \alpha }=\frac{\tan \alpha -m_2}{1+m_2\tan \alpha }{m}_1+m_1{m}_2\tan \alpha -\tan \alpha -m_2{\tan }^2\alpha =\tan \alpha -m_2+m_1{\tan }^2\alpha -m_1{m}_2\tan \alpha m_1+m_2-(m_1+m_2){\tan }^2\alpha -2\tan \alpha +2m_1{m}_2\tan \alpha =0(1-{\tan }^2\alpha )(m_1+m_2)-2\tan \alpha +2\tan \alpha m_1{m}_2=0(m_1+m_2)-\frac{2\tan \alpha }{1-{\tan }^2\alpha }+\frac{2\tan \alpha }{1-{\tan }^2\alpha }{m}_1{m}_2=0(m_1+m_2)-\tan 2\alpha +\tan 2\alpha m_1{m}_2=0$

using $\left(i\right)$ and $\left(ii\right)$

$\frac{k}{h}-\tan 2\alpha +\tan 2\alpha \frac{a}{h}=0k-h\tan 2\alpha +a\tan 2\alpha =0k=h\tan 2\alpha -a\tan 2\alpha$

Replacing $h$ by $x$ and $k$ by $y$

$y=\tan 2\alpha (x-a)$