The equation of the tangent to the curve $y=\pi e^{\frac{-x}{\pi}}$ at the point where it crosses the $y$-axis is

The tangent to a non-linear curve $y = f(x)$ at any point $P(x, y)$ intersects the $x$-axis and $y$-axis at $A$ and $B$ respectively. If the normal to the curve $y = f(x)$ at $P$ intersects the $y$-axis at $C$ such that $AC = BC$,and $f(2) = 3$,then the equation of the curve is:

$p_1$ and $p_2$ are the perpendicular distances from the origin to the tangent and normal drawn at any point on the curve $x^{2/3} + y^{2/3} = a^{2/3}$ respectively. If $k_1 p_1^2 + k_2 p_2^2 = a^2$,then $k_1 + k_2 =$

If the angle between the curves $y=e^{2(1+x)-4}$ and $x^2 y=1$ at the point $(1,1)$ is $\theta$,then $|\sin \theta|+|\cos \theta|=$

If the normal to the curve $y = f(x)$ is parallel to the $x-$axis,then the correct statement is:

Show that the normal at any point $\theta$ to the curve $x=a \cos \theta+a \theta \sin \theta, y=a \sin \theta-a \theta \cos \theta$ is at a constant distance from the origin.