Question

The tangent to the curve $y=a{x}^2+bx$ at $\left(2,-8\right)$ 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

$a=2,b=-8$

Explanation for the correct answer:

Given curve is, $y=a{x}^2+bx$

The slope of the tangent to a curve at each point is the derivative of the curve.

Thus, the slope of the tangent is described as,

$\begin{array}{rcl}\frac{\mathrm{d}y}{\mathrm{d}x}& =& \frac{\mathrm{d}}{\mathrm{d}x}\left(a{x}^2+bx\right)\\ & =& 2ax+b\end{array}$

Given, the tangent is parallel to $X$-axis at $\left(2,-8\right)$.

The slope of the tangent at this point is,

$\begin{array}{rcl}{\left(\frac{dy}{dx}\right)}_{\left(2,-8\right)}& =& 2a·2+b\\ & =& 4a+b\end{array}$

Since the tangent is parallel to the $x$-axis, its slope is $0$ (because the slope of $x$-axis is $0$ and parallel lines have equal slope).

$\begin{array}{cc}\Rightarrow & 0=4a+b\\ \Rightarrow & b=-4a\end{array}$

We know that the point $\left(2,-8\right)$ is on the given curve. This means that the point satisfies the equation of the curve.

$\begin{array}{ccc}\Rightarrow & -8=a\times 2^2+b\times 2& \\ \Rightarrow & -8=4a-4a\times 2& \left[\because b=-4a\right]\\ \Rightarrow & -8=-4a& \\ \Rightarrow & a=2& \end{array}$

But,

$\begin{array}{rcl}& & \begin{array}{cc}b=-4a& \\ =-4\times 2& \left[\because a=2\right]\\ =-8& \end{array}\end{array}$

Therefore, $a=2$ and $b=-8$

Hence, option C is correct.