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 $y=x^2+6x+10$. Find the value of $2(a-b).$

• A. 5
• B. 7
• C. 6
• D. 8

Answer 1

The correct option is B 7

Given curve is $y=a{x}^2+bx+\frac{7}{2}$
Differentiating:
$\frac{dy}{dx}=2ax+b$
${\left(\frac{dy}{dx}\right)}_{(1,2)}=2a+b$
$\Rightarrow m_1=2a+b$
Also, since $(1,2)$ lies on this curve, $2=a+b+\frac{7}{2}$ ... (1)
Another curve given is $y=x^2+6x+10$
Differentiating:
$\frac{dy}{dx}=2x+6$
${\left(\frac{dy}{dx}\right)}_{(-2,2)}=2$
$\Rightarrow$ Slope of the normal is $m_2=\frac{-1}{2}$
Since, the tangent and normal are parallel, $m_1=m_2$
$\Rightarrow 2a+b=\frac{-1}{2}$ ... (2)
Solving (1) and (2), we get $a=1$ and $b=\frac{-5}{2}$
Hence, the value of $2(a-b)=2\times \frac{7}{2}=7$