The equation of straight line which passes through the point \(\left(a \cos ^{3} \theta, a \sin ^{3} \theta\right)\) and perpendicular to \(x \sec \theta+y \operatorname{cosec} \theta=a\) is

Given,
Point \(\left(x_{1}, y_{1}\right)=\left(a \cos ^{3} \theta, a \sin ^{3} \theta\right)\)
Other line \(\Rightarrow x \sec \theta+y \operatorname{cosec} \theta=a\)
\(\begin{aligned}
y &=-\frac{x \sec \theta}{\operatorname{cosec} \theta}+\frac{a}{\operatorname{cosec} \theta} \\
\Rightarrow \quad y &=\left(-\frac{\sin \theta}{\cos \theta}\right) x+a \sin \theta \\
\Rightarrow \quad y &=m_{1} x+c \\
\therefore \quad m_{1} &=-\frac{\sin \theta}{\cos \theta}
\end{aligned}\)
Equation of required line
\(\begin{aligned}
&\Rightarrow\left(y-y_{1}\right)=m\left(x-x_{1}\right) \\
&\text { where } m=\frac{\cos \theta}{\sin \theta}\left[\because m_{1} \cdot m=-1, m=-\frac{1}{m_{1}}\right.
\end{aligned}\left.=-\frac{1}{\left(-\frac{\sin \theta}{\cos \theta}\right)}=\frac{\cos \theta}{\sin \theta}\right]\)
\(\left(y-a \sin ^{3} \theta\right)=\frac{\cos \theta}{\sin \theta}\left(x-a \cos ^{3} \theta\right)\)
\(\Rightarrow \quad y \sin \theta-a \sin ^{4} \theta=x \cos \theta-a \cos ^{4} \theta\)
\(\Rightarrow \quad x \cos \theta-y \sin \theta=a\left(\cos ^{4} \theta-\sin ^{4} \theta\right)\)
\(\quad=a\left(\cos ^{2} \theta+\sin ^{2} \theta\right)\left(\cos ^{2} \theta-\sin ^{2} \theta\right)\)
\(\Rightarrow \quad x \cdot(1) \cdot(\cos 2 \theta)=a \cos 2 \theta\)
\(\quad x \cos \theta-y \sin \theta=a \cos 2 \theta\)