Find the equation of all lines having slope $2$ which are tangents to the curve $y = \frac{1}{x-3}, x \neq 3$.

Let $f: R \rightarrow R$ be a bijection. $A$ curve represented by $y=f(x)$ is such that $f^{\prime}(x)>0$ for all $x \in R$. The tangent and normal drawn at $P(\alpha, 1)$ on the curve cut the $X$-axis at $A$ and $B$ respectively,and $C$ is the foot of the perpendicular from $P$ onto the $X$-axis. If $P(\alpha, 1)$ is such a point that $AC+CB$ is minimum,then the tangent at $P$ is parallel to the line

The lengths of tangent,subtangent,normal and subnormal for the curve $y=x^2+x-1$ at $(1,1)$ are $A, B, C$ and $D$ respectively,then their increasing order is

The curve $y=ax^3+bx^2+cx+5$ touches the $X$-axis at $(-2,0)$ and cuts the $Y$-axis at a point $Q$ where its gradient is $3$. Then the values of $a, b, c$ respectively are:

If the straight line $x \cos \alpha + y \sin \alpha = p$ touches the curve $(\frac{x}{a})^n + (\frac{y}{b})^n = 2$ at the point $(a, b)$ on it and $\frac{1}{a^2} + \frac{1}{b^2} = \frac{k}{p^2}$,then $k =$

$A$ 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