Find the slope of the normal to the curve $x = a \cos^3 \theta$,$y = a \sin^3 \theta$ at $\theta = \pi / 4$.

$A$ curve is given by the equations $x = a \cos \theta + \frac{1}{2}b \cos 2\theta$ and $y = a \sin \theta + \frac{1}{2}b \sin 2\theta$. The points for which $\frac{d^2y}{dx^2} = 0$ are given by:

If $x=5(1-\sin t)$ and $y=5(t+\cos t)$, then $\frac{dx}{dy}=$

If $x = \sec \theta - \cos \theta$ and $y = \sec^n \theta - \cos^n \theta$,then $\left(\frac{dy}{dx}\right)^2$ is equal to

$x=\cos ^{-1}\left(\frac{1}{\sqrt{1+t^2}}\right), y=\sin ^{-1}\left(\frac{t}{\sqrt{1+t^2}}\right) \Rightarrow \frac{d y}{d x}$ is equal to

If $x$ and $y$ are connected parametrically by the equations,without eliminating the parameter,find $\frac{dy}{dx}$ for $x=a\left(\cos t+\log \tan \frac{t}{2}\right), y=a \sin t$.