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.

If the curves $2x = y^2$ and $2xy = K$ intersect perpendicularly,then the value of $K^2$ is

The coordinates of the point $P$ on the curve $x=a(\theta+\sin \theta), y=a(1-\cos \theta)$, where the tangent is inclined at an angle $\frac{\pi}{4}$ to the $x$-axis, are

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 slope of the tangent to the curve $y = \int_0^x \frac{1}{1+t^3} dt$ at the point where $x = 1$ is

Let $a, b, c \in \mathbb{R}$ be such that $2a + 3b + 6c = 0$ and $g(x)$ be the antiderivative of $f(x) = ax^2 + bx + c$. If the slopes of the tangents drawn to the curve $y = g(x)$ at $(1, g(1))$ and $(2, g(2))$ are equal, then