If the normal to the curve y(x)=_0^x(2t^2-15t+10),dt at a point (a,b) is parallel to the l…

Question

If the normal to the curve $y(x)=\displaystyle\int_0^x(2t^2-15t+10)\,dt$ at a point $(a,b)$ is parallel to the line $x+3y=-5$, $a>1$, then the value of $|a+6b|$ is equal to.........

Answer: 406

Accepted Answer

$y(x)=\displaystyle\int_0^x(2t^2-15t+10)dt\Rightarrow y'(x)=2x^2-15x+10$

Since normal is parallel to $x+3y=-5$ (slope $-1/3$), slope of tangent at $(a,b)$ equals slope of the given line:
$$y'(a)=3\Rightarrow2a^2-15a+10=3\Rightarrow2a^2-15a+7=0\Rightarrow(2a-1)(a-7)=0\Rightarrow a=\frac12\text{ or }7$$

Since $a>1$, $a=7$. Then
$$b=\int_0^7(2t^2-15t+10)dt=\frac{2}{3}(7)^3-\frac{15}{2}(7)^2+10(7)=-\frac{413}{6}$$

So $a+6b=7-413=-406\ \Rightarrow\ |a+6b|=406$

Solutions · 1