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Question

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

A

$a=2,b=-2$

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B

$a=2,b=-4$

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C

$a=2,b=-8$

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D

$a=4,b=-4$

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Solution

## 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).$\begin{array}{cc}⇒& 0=4a+b\\ ⇒& b=-4a\end{array}$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.$\begin{array}{ccc}⇒& -8=a×{2}^{2}+b×2& \\ ⇒& -8=4a-4a×2& \left[\because b=-4a\right]\\ ⇒& -8=-4a& \\ ⇒& a=2& \end{array}$But,$\begin{array}{rcl}& & \begin{array}{cc}b=-4a& \\ =-4×2& \left[\because a=2\right]\\ =-8& \end{array}\end{array}$Therefore, $a=2$ and $b=-8$Hence, option C is correct.

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