Find the slope of the tangent to the curve $y = x^{3} - 3x + 2$ at the point whose $x$-coordinate is $3$.

The angle between the curves $y^2=x$ and $x^2=y$ at the origin is:

The coordinates of the point$(s)$ on the graph of the function $f(x) = \frac{x^3}{3} - \frac{5x^2}{2} + 7x - 4$ where the tangent drawn cuts off intercepts from the coordinate axes which are equal in magnitude but opposite in sign,are:

If the curves $\frac{x^2}{a^2} + \frac{y^2}{4} = 1$ and $y^3 = 16x$ intersect at right angles,then $a^2 = \dots$

If the slope of the line through $(0,0)$ which is tangent to the curve $y=x^2+x+16$ is $m$,then the value of $m-4$ is

The length of the normal to the curve $x = a(\theta + \sin \theta)$,$y = a(1 - \cos \theta)$ at the point $\theta = \pi/2$ is