We have the curve:
[tex]y^2-2xy=16[/tex]We have to find dy/dx of this curve and the equation of the horizontal tangent line to the curve.
We can find the first derivative dy/dx using implicit differentiation:
[tex]\begin{gathered} \frac{d}{dx}(y^2-2xy)=\frac{d}{dx}(16) \\ 2y*\frac{dy}{dx}-2\frac{d}{dx}(xy)=0 \\ 2y*\frac{dy}{dx}-2(x*\frac{dy}{dx}+1*y)=0 \\ 2y*\frac{dy}{dx}-2x*\frac{dy}{dx}-2y=0 \\ 2(y-x)*\frac{dy}{dx}-2y=0 \\ (y-x)*\frac{dy}{dx}-y=0 \\ (y-x)*\frac{dy}{dx}=y \\ \frac{dy}{dx}=\frac{y}{y-x} \end{gathered}[/tex]The value of dy/dx will represent the slope of the tangent line to the point (x,y).
Horizontal lines have slopes m = 0, so we can find the relation between x and y as:
[tex]\begin{gathered} \frac{dy}{dx}=0 \\ \frac{y}{y-x}=0 \\ y=0 \end{gathered}[/tex]In this case, we don't have a defined value of x. We can see that from the graph:
The curve is an hyperbola where y = 0 is one of the asymptotes.
Then, we can consider y = 0 as the tangent line on the infinity.
Answer: the tangent line to the curve is y = 0.
