Question

If tangent at point $(1,2)$ on the curve $y=a{x}^2+bx+\frac{7}{2}$ is parallel to normal at $(-2,2)$ on the curve $y=x^2+6x+10,$ then

• A. $a=1$
• B. $a=-1$
• C. $b=-\frac{15}{2}$
• D. $b=\frac{5}{2}$

Answer 1

The correct option is A $a=1$

$y=x^2+6x+10$
${\frac{dy}{dx}|}_{(-2,2)}=-4+6=2$
Slope of normal is $-\frac{1}{2}.$

Curve $y=a{x}^2+bx+\frac{7}{2}$ passes through $(1,2)$
$\Rightarrow 2=a+b+\frac{7}{2}\dots \left(1\right)$

${\frac{dy}{dx}|}_{(1,2)}=2a+b$
Since tangent is parallel to normal,
$\therefore -\frac{1}{2}=2a+b\dots \left(2\right)$
Solving $\left(1\right)$ and $\left(2\right),$
$a=1,b=\frac{-5}{2}$