If the normal to the curve $y(x)=\int_{0}^{x}(2t^{2}-15t+10)dt$ at a point $(a, b)$ is parallel to the line $x+3y=-5$,where $a>1$,then the value of $|a+6b|$ is equal to..........

The sum of the intercepts made by a tangent drawn to the curve $\left(\frac{x}{a}\right)^n + \left(\frac{y}{b}\right)^n = 2$ at the point $(a, b)$ on the coordinate axes 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.

The angle between the curves $y^2=x$ and $x^2=y$ at the origin is:

Find the point on the curve $y = \sqrt{4x - 3} - 1$ where the tangent has a slope of $\frac{2}{3}$.

Find the equations of the tangent and the normal to the curve $y=x^{4}-6 x^{3}+13 x^{2}-10 x+5$ at the point $(1,3)$.