Question

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

Answer 1

First we find the derivative of xy= 16.

Using implicit differentiation, we have

`\begin{gathered} xy=16 \\ x(dy)/(dx)+y=0 \\ x(dy)/(dx)=-y \\ (dy)/(dx)=-(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 }(dy)/(dx)=-4 \end{gathered}`

So we have

`\begin{gathered} -(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=(16)/(4) \\ x^2=4 \\ x=\pm√(4) \\ x=\pm2 \end{gathered}`

Hence the answer is option C