Question

The tangent to $y=a{x}^2+bx+\frac{7}{2}$ at $(1,2)$ is parallel to the normal at the point $(-2,2)$ on the curve $y=x^2+6x+10$ Then the value of $(a+b)$ is_____.

Answer 1

Answer: -1.5

$y=a{x}^2+bx+\frac{7}{2}$
$\Rightarrow \frac{dy}{dx}=2ax+b$
$(1,2)$ lies on the curve
$\Rightarrow 2=a+b+\frac{7}{2}$
$\Rightarrow a+b=-\frac{3}{2}\dots (i)$
${\frac{dy}{dx}|}_{(1,2)}=2a+b$ (slope of tangent)
For $y=x^2+6x+10,\frac{dy}{dx}=2x+6$
${\frac{dy}{dx}|}_{(-2,2)}=2$
$\therefore$ slope of normal $=-\frac{1}{2}$
then $2a+b=-\frac{1}{2}\dots (ii)$
using (i) and (ii), $a=1$ and $b=-\frac{5}{2}$