Question

Three normals are drawn to the parabola $y^2=4ax$ cos a from any point lying on the straight line $y=bsina$. Prove that the locus of the orthocentre of the triangles formed by the corresponding tangents is the curve $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, angle a being variable.

Answer 1

The normal to the parabola $y^2=4ax$ is $y=mx-2am-a{m}^3$

Since, here $a$ given as $a\cos a$

And the normal passes through a line, $y=b\sin a$

$b\sin a=mx-2am\cos a-a\cos a{m}^3$

$m^{3}a\cos a+m(2a\cos a-x)+b\sin a=0$

The above equation is cubic in $m,$ we get

$m_1+m_2+m_3=0$

$m_1{m}_2+m_2{m}_3+m_1{m}_3=\frac{2a\cos a-x}{a\cos a}$

$m_1{m}_2{m}_3=\frac{-b\sin a}{a\cos a}$

Let the locus of the orthocentre of the triangles formed by the corresponding tangents is $(h,k)$

The tangent to the parabola $y^2=4ax\cos a$ is $y=mx+\frac{a\cos a}{m}$

So, $T_1=y=m_{1}x+\frac{a\cos a}{m_1}$

Similarly we will get $T_2,T_3$

In triangle $ABC$

The intersection point of $T_1,T_2$ is $x=\frac{a\cos a}{m_1{m}_2}$ , $y=a\cos a(\frac{1}{m_1}+\frac{1}{m_2})$

The slope of $T_3$ is $m_3$

The perpendicular slope is $\frac{-1}{m_3}$

Therefore the equation of altitude $AD$ is

$y-a\cos a(\frac{1}{m_1}+\frac{1}{m_2})=(x-\frac{a\cos a}{m_1{m}_2})\left(\frac{-1}{m_3}\right)$

Similarly other altitude can be found as $BE$ as

$y-a\cos a(\frac{1}{m_3}+\frac{1}{m_2})=(x-\frac{a\cos a}{m_3{m}_2})\left(\frac{-1}{m_1}\right)$

The intersection point of these two altitudes is our required orthocentre,

As we know $y=k=b\sin a$

and $x$ can be found from solving the two altitude equations as,

$x=h=-a\cos a$

$\frac{k}{b}=\sin a$

$\frac{h}{a}=-\cos a$

Squaring on both sides, we get

$\frac{h^2}{a^2}+\frac{k^2}{b^2}=1$

Genaralising the equation, we get

$\frac{x^2}{a^2}+\frac{x^2}{b^2}=1$