Find the slope of the tangent to the curve $y=3x^{4}-4x$ at $x=4$.

The tangent to the curve $y=e^x$ drawn at the point $(c, e^c)$ intersects the line joining the points $(c-1, e^{c-1})$ and $(c+1, e^{c+1})$:

If the length of the subnormal at any point on the curve $y = a^{1-n}x^n$ is constant,then $n = \dots$

If the normal drawn at a point $P$ on the curve $3y = 6x - 5x^3$ passes through $(0,0)$,then the positive integral value of the abscissa of the point $P$ is

If all the normals drawn to the curve $y=\frac{1+3x^2}{3+x^2}$ at the points of intersection of $y=\frac{1+3x^2}{3+x^2}$ and $y=1$ pass through the point $(\alpha, \beta)$,then $3\alpha+2\beta=$

$P$ is the point of contact of the tangent drawn from the origin to the curve $y = \log_{e} x$. The length of the perpendicular drawn from the origin to the normal at $P$ is