The equation of the tangent to the curve $x = \frac{t - 1}{t + 1}, y = \frac{t + 1}{t - 1}$ at $t = 2$ is:

The function $y(x)$ represented by $x=\sin t$,$y=a e^{t \sqrt{2}}+b e^{-t \sqrt{2}}$,$t \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)$ satisfies the equation $(1-x^2) y^{\prime \prime}-x y^{\prime}=k y$,then the value of $k$ is

If $x = a \sin 2\theta (1 + \cos 2\theta )$ and $y = b \cos 2\theta (1 - \cos 2\theta )$,then $\frac{dy}{dx} = $

The position of a point at time $t$ is given by $x = a + bt - ct^2$ and $y = at + bt^2$. The resultant acceleration of the point at time $t$ is given by:

At any two points of the curve represented parametrically by $x = a(2 \cos t - \cos 2t)$ and $y = a(2 \sin t - \sin 2t)$,the tangents are parallel to the $x$-axis. The values of the parameter $t$ corresponding to these points differ from each other by:

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