At what point on the curve ${x^3} - 8{a^2}y = 0$ is the slope of the normal equal to $\frac{-2}{3}$?

The normal to the curve $2x^2 + y^2 = 12$ at the point $(2, 2)$ meets the curve again at

The equation of the tangent to the curve $(1 + x^2)y = 2 - x$ at the point where it crosses the $x$-axis is:

Consider a curve $y=y(x)$ in the first quadrant as shown in the figure. Let the area $A_{1}$ be twice the area $A_{2}$. Then the normal to the curve perpendicular to the line $2x - 12y = 15$ does $NOT$ pass through the point.

$A(-2,9)$ and $B(1,6)$ are two points on the curve $y=x^2+5$. The coordinates of the point $C$ on the curve such that the tangent drawn at $A$ is parallel to the chord $BC$,are

For $h, k \in N$,let $P(h, k)$ be the point of intersection of the curves $x^2 y - x^3 = 8$ and $y^3 - x y^2 = 32$. If $\theta$ is the acute angle between these two curves at $P$,then $\tan \theta =$