Question

Prove that the locus of the feet of the perpendiculars drawn from the vertex of the parabola upon chords, which subtend an angle of ${45}^o$ at the vertex, is the curve
$r^2-24ar\cos \theta +16a^2\cos 2\theta =0$.

Answer 1

Let the feet of perpendicular from vertex be $P(r\cos \theta ,r\sin \theta )$

Slope of perpendicular $=m=\tan \theta$

$\Rightarrow$ slope of chord $=-\cot \theta$

Equation of chord through $P$ is

$y-r\sin \theta =-\cot \theta (x-r\cos \theta )(x-r\cos \theta )=-\tan \theta (y-r\sin \theta )(x-r\cos \theta )=-\frac{\sin \theta }{\cos \theta }(y-r\sin \theta )x\cos \theta -r{\cos }^2\theta =-y\sin \theta +r{\sin }^2\theta x\cos \theta +y\sin \theta =r\frac{x\cos \theta +y\sin \theta }{r}=\mathrm{1.........}\left(i\right)$

Making the equation of parabola homogenous using $\left(i\right)$

$y^2=4ax{y}^2=4ax\left(\frac{x\cos \theta +y\sin \theta }{r}\right)r{y}^2=4a{x}^2\cos \theta +4axy\sin \theta 4a{x}^2\cos \theta +4axy\sin \theta -r{y}^2=0$

Now $\tan \theta =\frac{2\sqrt{h^2-ab}}{a+b}$

$\tan \theta =\frac{2\sqrt{h^2-ab}}{a+b}\tan 45=\frac{2\sqrt{h^2-ab}}{a+b}1=\frac{2\sqrt{h^2-ab}}{a+b}a+b=2\sqrt{h^2-ab}{a}^2+b^2+2ab=4h^2-4ab{a}^2+b^2-4h^2+6ab=0{\left(4a\cos \theta \right)}^2+r^2-4{\left(2a\sin \theta \right)}^2+6\left(4a\cos \theta \right)(-r)=016a^2{\cos }^2\theta +r^2-16a^2{\sin }^2\theta -24a\cos \theta r=0r^2-24a\cos \theta r+16a^2({\cos }^2\theta -{\sin }^2\theta )=0r^2-24a\cos \theta r+16a^2\cos 2\theta =0$