The equation of the line, passing through the point $\left(a \cos ^3 \theta, a \sin ^3 \theta\right)$ and…
The equation of the line, passing through the point $\left(a \cos ^3 \theta, a \sin ^3 \theta\right)$ and perpendicular to the line $x \cos$ $\theta-y \sin \theta=a$, is
$2 x \sin \theta+2 y \cos \theta=a \sin 2 \theta$
$x \cos \theta-y \sin \theta=a \sin 2 \theta$
$x \sin \theta+y \sin \theta=a \cos 2 \theta$
$x \sin \theta-y \cos \theta=a \cos 2 \theta$
Solution
Slope of line $\mathrm{x} \cos \theta-\mathrm{y} \sin \theta=\mathrm{a}$ is
$\mathrm{m}^{\prime}=\cot \theta$
So, the slope of the required line is:
$m=\frac{-1}{m^{\prime}}=-\tan \theta$
Now, the required equation of line is:
$\left(y-a \sin ^3 \theta\right)=m\left(x-a \cos ^3 \theta\right)$
$\begin{aligned} & \Rightarrow y-a \sin ^3 \theta=-\tan \theta\left(x-a \cos ^3 \theta\right) \\ & \Rightarrow y-a \sin ^3 \theta=-x \tan \theta+a \cos ^3 \theta \tan \theta \\ & \Rightarrow y+x \tan \theta=a \sin ^3 \theta+a \cos ^2 \theta \sin \theta \\ & \Rightarrow y+x \frac{\sin \theta}{\cos \theta}=a \sin \theta\left(\sin ^2 \theta+\cos ^2 \theta\right) \\ & \Rightarrow y \cos \theta+x \sin \theta=a \sin \theta \cdot \cos \theta \\ & \Rightarrow 2 x \sin \theta+2 y \cos \theta=a \sin 2 \theta\end{aligned}$