Let $C$ be a curve given by $y(x)=1+\sqrt{4x-3}$,$x>\frac{3}{4}$. If $P$ is a point on $C$ such that the tangent at $P$ has a slope of $\frac{2}{3}$,then a point through which the normal at $P$ passes is:

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):

At which points on the curve $y = x^3 - 3x^2 - 9x + 7$ is the tangent parallel to the $x$-axis?

What is the angle of intersection of the curves $y = 4 - x^2$ and $y = x^2$?

If $P$ and $Q$ are two different points on the curve $y = x^3 - x$ such that the tangent at $P$ intersects the curve again at $Q$,then $\frac{m_{OQ} + 1}{m_{OP} + 1}$ is equal to,where $O$ is the origin and $m_{AB}$ represents the slope of the line segment $AB$.

If the normal to the curve $y = f(x)$ at the point $(4, 6)$ makes an angle $\frac{2\pi}{3}$ with the positive $x$-axis in the anticlockwise direction,then $f'(4)$ is: