Find the equation of the normal to the curve $2y = 3 - x^2$ at the point $(1, 1)$.

$P(5,2)$ is a point on the curve $y=f(x)$ and $\frac{7}{2}$ is the slope of the tangent to the curve at $P$. The area of the triangle (in sq. units) formed by the tangent and the normal to the curve at $P$ with the $x$-axis is:

The slope of the normal drawn at a point $P$ to the curve $y = x^3 - 10x^2 + 31x - 30$ is $-\frac{1}{14}$. Find the $x$-intercept of the tangent at point $P$, given that the $x$-coordinate of $P$ is an integer.

If the equation of the normal to the curve $x^3 - y^2 = 0$ at the point $(m^2, -m^3)$ is $y = 3mx - 4m^3$,then $m^2 = \dots\dots$.

The equation of the tangent at $x = \frac{\pi}{2}$ to the curve $y = x \sin x$ is

If the equation of a tangent drawn to the curve $y=\cos (x+y), -1 \leq x \leq 1+\pi$ is $x+2y=k$, then $k=$