# If the normal to the curve

Question:

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

Solution:

$y(x)=\int_{0}^{x}\left(2 t^{2}-15 t+10\right) d t$

$\left.y^{\prime}(x)\right]_{x=a}=\left[2 x^{2}-15 x+10\right]_{a}=2 a^{2}-15 a+10$

Slope of normal $=-\frac{1}{3}$

$\Rightarrow \quad 2 a^{2}-15 a+10=3 \Rightarrow a=7$

$\& \quad a=\frac{1}{2}$ (rejected)

$b=y(7)=\int_{0}^{7}\left(2 t^{2}-15 t+10\right) d t$

$=\left[\frac{2 t^{3}}{3}-\frac{15 t^{2}}{2}+10 t\right]_{0}^{7}$

$\Rightarrow \quad 6 b=4 \times 7^{3}-45 \times 49+60 \times 7$

$|a+6 b|=406$