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 $| a +6 b |$ is equal to ________.

Answer 1

$y ( x )=\int_{0}^{ x }\left(2 t ^{2}-15 t +10\right) dt$
$\left.y ^{\prime}( x )\right]_{ x = a }=\left[2 x ^{2}-15 x +10\right]_{ a }=2 a ^{2}-15 a +10$
Slope of normal $=-\frac{1}{3}$
$\Rightarrow 2 a ^{2}-15 a +10=3 \Rightarrow a =7$
$\& a =\frac{1}{2} $ (rejected)
$b = y (7)=\int_{0}^{7}\left(2 t ^{2}-15 t +10\right) dt$
$=\left[\frac{2 t ^{3}}{3}-\frac{15 t ^{2}}{2}+10 t \right]_{0}^{7}$
$\Rightarrow 6 b =4 \times 7^{3}-45 \times 49+60 \times 7$
$| a +6 b |=406$