Question

The tangent to the curve $y=a{x}^2+bx$ at (2, -8) is parallel to x-axis. Then

• A. a=2, b=-2
• B. a=2, b=-4
• C. a=2 b=-8
• D. a=4, b=-4

Answer 1

The correct option is C a=2 b=-8

$y=a{x}^2+bx\frac{dy}{dx}=2ax+b\Rightarrow {\left(\frac{dy}{dx}\right)}_{(2,-8)}=4a+b$
$\because$ Tangent is parallel to x-axis
$\therefore \frac{dy}{dx}=0\Rightarrow b=-4a....\left(i\right)$
Now, point (2, –8) is on the curve of $y=a{x}^2+bx$
$\therefore -8=4a+2b......\left(ii\right)$
From (i) and (ii), we get a =2, b=-8.