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 $y=x^{2}+6 x+10$ Then the value of $(a+b)$ is_____.

Answer 1

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