Find the equation of the line having slope $0$ which is tangent to the curve $y=\frac{1}{x^{2}-2x+3}$.

The point on the curve $y=x^3$,at which the tangent to the curve is parallel to the $X$-axis,is

The curve $\left(\frac{x}{a}\right)^n+\left(\frac{y}{b}\right)^n=2$,where $n \in N$,touches the line at the point $(a, b)$. Find the equation of the line.

The area of the triangle formed by the tangent and the normal drawn to the curve $y^2=4x$ at $(1,2)$ with the $Y$-axis is (in square units):

The number of those tangents to the curve $y^2 - 2x^3 - 4y + 8 = 0$ which pass through the point $(1, 2)$ is

If $A = \{P(\alpha, \beta) \mid \text{the tangent drawn at } P \text{ to the curve } y^3 - 3xy + 2 = 0 \text{ is a horizontal line}\}$ and $B = \{Q(a, b) \mid \text{the tangent drawn at } Q \text{ to the curve } y^3 - 3xy + 2 = 0 \text{ is a vertical line}\}$, then $n(A) + n(B) = $