Consider $f(x) = \tan^{-1}\left(\sqrt{\frac{1 + \sin x}{1 - \sin x}}\right)$,where $x \in (0, \frac{\pi}{2})$. $A$ normal to $y = f(x)$ at $x = \frac{\pi}{6}$ also passes through the point:

The number of points on the curve $y=54x^5-135x^4-70x^3+180x^2+210x$ at which the normal lines are parallel to $x+90y+2=0$ is:

If $f: R \rightarrow R$ is a function defined for all $x \in R$ by $f(x)=x^3+f^{\prime}(1) x^2+f^{\prime \prime}(2) x-f^{\prime \prime \prime}(3)$,then the area (in sq. units) of the triangle formed by the $X$-axis,the tangent,and the normal drawn to the curve $y=f(x)$ at $x=0$ is

Find the points on the curve $\frac{x^{2}}{4} + \frac{y^{2}}{25} = 1$ at which the tangents are parallel to the $x$-axis.

Prove that the curves $x=y^{2}$ and $xy=k$ cut at right angles if $8k^{2}=1$.

The two curves $x=y^2$ and $xy=a^3$ cut orthogonally at a point,then $a^2$ is equal to