The tangent at the point $(2, -2)$ to the curve $x^2y^2 - 2x = 4(1 - y)$ does not pass through the point:

The points on the curve $y^2 = x + \sin x$ at which the normal is parallel to the $Y$-axis lie on

If the tangent to the curve $y = \frac{x}{x^2-3}$,$x \in R, (x \neq \pm \sqrt{3})$ at a point $(\alpha, \beta) \neq (0,0)$ on it,is parallel to the line $2x + 6y - 11 = 0$,then

For the curve $\frac{x^n}{a^n}+\frac{y^n}{b^n}=2, (n \in N \text{ and } n > 1)$,the line $\frac{x}{a}+\frac{y}{b}=2$ is

The area (in sq units) of the triangle formed by the normal drawn at the point $(1,0)$ on the curve $x=e^{\sin y}$ with the coordinate axes is

The value of $c$ such that the straight line joining the points $(0,3)$ and $(5,-2)$ is tangent to the curve $y=\frac{c}{x+1}$ is