Question

Through the vertex A of the parabola $y^2=4ax$ two chords AP and AQ are drawn, and the circles on AP and AQ as diameters intersect in R. Prove that, if ${\theta }_1,{\theta }_2$ and $\varphi$ be the angles made with the axis by the tangents at P and Q and by AR, then
$\cot {\theta }_1+\cot {\theta }_2+2\tan \varphi =0.$

Answer 1

Vertex of the parabola is $A(0,0)$

Let the point $P$ and $Q$ be $(a{t}_1^2,2a{t}_1)$ amd $(a{t}_2^2,2a{t}_2)$

Equation of circle with $AP$ as diameter is

$(x-0)(x-a{t}_1^2)+(y-0)(y-2a{t}_1)=0x^2+y^2-a{t}_1^{2}x-2a{t}_{1}y=\mathrm{0.......}\left(i\right)$

Similarly equation of circle with $AQ$ as diameter is

$x^2+y^2-a{t}_2^{2}x-2a{t}_{2}y=\mathrm{0........}\left(ii\right)$

Equation of common chord of circle is obtained by subtracting equation of both circles.

So the equation of common chord is $\left(i\right)-\left(ii\right)$

$x^2+y^2-a{t}_1^{2}x-2a{t}_{1}y-x^2-y^2+a{t}_2^{2}x+2a{t}_{2}y=0(t_2^2-t_1^2)x+2(t_2-t_1)y=0(t_2+t_1)x+2y=0y=-\frac{(t_2+t_1)}{2}x$

$\tan \varphi =$ slope of $AR$

$\tan \varphi =-\frac{(t_2+t_1)}{2}2\tan \varphi =-(t_2+t_1).....\left(i\right)$

Equation of tangent at $P$ is

$t_{1}y=x+a{t}_1^{2}x-t_{1}y+a{t}_1^2=0$

Slope of tangent at $P$ $=\frac{1}{t_1}$

$\tan {\theta }_1=\frac{1}{t_1}\cot {\theta }_1=t_1......\left(ii\right)$

Similarly slope of tangent at $Q$ is $\frac{1}{t_2}$

$\tan {\theta }_2=\frac{1}{t_2}\cot {\theta }_2=t_2......\left(iii\right)$

Adding $\left(i\right),\left(ii\right)$ and $\left(iii\right)$

$\cot {\theta }_1+\cot {\theta }_2+2\tan \varphi =t_1+t_2-(t_2+t_1)\cot {\theta }_1+\cot {\theta }_2+2\tan \varphi =0$

Hence proved