Question

(calculus !) at what values of X on the curve xy= 16 is the tangent line parallel to 4x + y = 0

Answer 1

SOLUTION

First we find the derivative of xy= 16.

Using implicit differentiation, we have

$\begin{gathered} xy=16 \\ x\frac{dy}{dx}+y=0 \\ x\frac{dy}{dx}=-y \\ \frac{dy}{dx}=-\frac{y}{x} \end{gathered}$

Writing 4x + y = 0 in slope intercept form we have

$\begin{gathered} 4x+y=0 \\ y=-4x \\ so\text{ the slope m, which is }\frac{dy}{dx}=-4 \end{gathered}$

So we have

$\begin{gathered} -\frac{y}{x}=-4 \\ -y=-4x \\ y=4x \end{gathered}$

We plug y = 4x into xy= 16, we have

$\begin{gathered} xy=16 \\ x(4x)=16 \\ 4x^2=16 \\ x^2=\frac{16}{4} \\ x^2=4 \\ x=\pm\sqrt{4} \\ x=\pm2 \end{gathered}$

Hence the answer is option C