Question

Show that three circles can be drawn to touch a parabola and also to touch at the focus a given straight line passing through the focus, and prove that the tangents at the point of contact with the parabola form an equilateral triangle.

Answer 1

We know that the equation to the circle which passes through the focus $(a,0)$ and also touches the oarabola $y^2=4ax$ at $(a{t}^2,2at)$ is $x^2+y^2-ax(3t^2+1)-ay(3t-t^3)+3a^2{t}^2=0$ and equation of any line is $x^2+y^2-ax(3t^2+1)-ay(3t-t^3)+3a^2{t}^2=0$ and equation of any line through $(a,0)$ is given by $y=\frac{\sin \alpha }{\cos \alpha }(x-a)$ $(\therefore m=\tan \alpha =\frac{\sin \alpha }{\cos \alpha })$

$x\sin \alpha -y\cos \alpha =a\sin \alpha$ and equation of perpendicular line will be

$x\cos \alpha +y\sin \alpha =a\cos \alpha$

Co-odinates of the centre of the circle given are $\left\{a\right(3t^2+1)/2,a(3t-t^3\left)/2\right\}$.

Since this passes through the centre then we have

$\frac{a(4t^2-1)}{2}\cos \alpha +\frac{a}{2}(3t-t^3)\sin \alpha =a\cos \alpha$

$\Rightarrow (3t^2+1)\cos \alpha +(3t-t^3)\sin \alpha =a\cos \alpha$

$\Rightarrow (3t-t^3)\sin \alpha =(1-3t^2)\cos \alpha$

$\therefore \cot \alpha =\frac{3t-t^3}{1-3t^2}$.

Putting $t=\tan \theta$ we get $\cot \alpha =\frac{3\tan \theta -{\tan }^3\theta }{1-3{\tan }^2\theta }=\tan 3\theta$

$\Rightarrow \tan 3\theta =\tan (\frac{\pi }{2}-\alpha )$

then we have

$3\theta -(\frac{\pi }{2}-\alpha ),\{\pi +(\frac{\pi }{2}-\alpha )\}$ or $\{2\pi +(\frac{\pi }{2}-\alpha )\}$

If ${\theta }_1,{\theta }_2$ and ${\theta }_3$ be three values of $\theta$, then

${\theta }_1=\frac{1}{3}(\frac{\pi }{2}-\alpha ),{\theta }_2=\frac{\pi }{3}+\frac{1}{3}(\frac{\pi }{2}-\alpha )$ and ${\theta }_3=\frac{2\pi }{3}+\frac{1}{3}(\frac{\pi }{2}-\alpha )$

So, ${\theta }_2-{\theta }_1=\frac{\pi }{3}={\theta }_3-{\theta }_2$.

therefore the three normal are inclined at ${60}^o$. So the angles between the tangents ar also ${60}^o$.

Therefore these form an equilateral triangle.