The equation of a line having slope $-\tan \theta$ and passing through the point $(a^2\sin \theta ,a^2\cos \theta )$ is

Slope $\left(m\right)=-\tan \theta$

Let $(x_1,y_1)=(a^2\sin \theta ,a^2\cos \theta )$

$y-y_1=m(x-x_1)$

$y-a^2\cos \theta =-\tan \theta (x-a^2\sin \theta )$

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

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

$x\sin \theta +y\cos \theta =a^2{\sin }^2\theta +a^2{\cos }^2\theta$

$x\sin \theta +y\cos \theta =a^2\left(1\right)$

$x\sin \theta +y\cos \theta =a^2$