What is the equation of the normal to the curve $y^2 = x^3$ at the point where the $x$-coordinate is $8$?

The equation of the normal to the curve $\sin y = \sqrt{3} x \sin \left(\frac{\pi}{6} + y\right)$ at $x = 0$ is:

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

Show that the line $\frac{x}{a} + \frac{y}{b} = 1$ touches the curve $y = b \cdot e^{-x / a}$ at the point where the curve intersects the $y$-axis.

If the normal to the curve $y = f(x)$ at the point $(3, 4)$ makes an angle of $3\pi / 4$ with the positive $X$-axis,find $f'(3)$.

Let $f(x)$ be a differentiable function,$A(0, \alpha)$ and $B(8, \beta)$ be two points on the curve $y=f(x)$. Given $f(0)=2$ and $f^{\prime}(4)=\frac{-3}{4}$. If the chord $AB$ of the curve is parallel to the tangent drawn at the point $(4, f(4))$,then $\beta=$