Find the points on the curve $\frac{x^{2}}{9}+\frac{y^{2}}{16}=1$ at which the tangents are parallel to the $y$-axis.

The slope of the tangent to the curve $(y-x^5)^2=x(1+x^2)^2$ at the point $(1,3)$ is

If $(a^2-1) x+a y+(3-a)=0$ is a normal to the curve $x y=1$,then the interval in which '$a$' lies is

If the points of contact of the tangents drawn from $(0,0)$ to the curve $y=x^2+3x+4$ are $(\alpha, \beta)$ and $(\gamma, \delta)$,then $\beta+\delta=$

The ratio between the length of the subtangent at any point other than the origin on the parabola $y^2 = 16ax$ and the abscissa of that point is:

$A$ curve is represented by the equations $x = \sec^2 t$ and $y = \cot t$,where $t$ is a parameter. If the tangent at the point $P$ on the curve where $t = \pi/4$ meets the curve again at the point $Q$,then the $x$-coordinate of $Q$ is equal to