The equation of a straight line which passes through the point a cos 3 θ , a sin 3 θ and…

We know that a line perpendicular to the line $ax+by+c=0$ is given by $bx-ay+k=0.$

Here, given line $x\sec \theta +ycosec\theta =a.$

So, the perpendicular line is $xcosec\theta -y\sec \theta +k=0.$

The point $\left(a{\cos }^3\theta ,a{\sin }^3\theta \right)$ lies on the line 

$\Rightarrow \frac{a{\cos }^3\theta }{\sin \theta }-\frac{a{\sin }^3\theta }{\cos \theta }=-k$

$\Rightarrow \frac{a\left({\cos }^4\theta -{\sin }^4\theta \right)}{\sin \theta \cos \theta }=-k$

$\Rightarrow \frac{a\left({\cos }^2\theta -{\sin }^2\theta \right)\left({\cos }^2\theta +{\sin }^2\theta \right)}{\sin \theta \cos \theta }=-k$

Using ${\cos }^2\theta +{\sin }^2\theta =1, {\cos }^2\theta -{\sin }^2\theta =\cos 2\theta ,$

$\Rightarrow \frac{a\cos 2\theta }{\sin \theta \cos \theta }=-k$

So, the required line is

$\frac{x}{\sin \theta }-\frac{y}{\cos \theta }=\frac{a\cos 2\theta }{\sin \theta \cos \theta }$

$\Rightarrow x\cos \theta -y\sin \theta =a\cos 2\theta .$