Question

Find dy/dx and write an equation of each horizontal tangent line to the curve. Y^2-2xy=16

Answer 1

We have the curve:

`y^2-2xy=16`

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:

`\begin{gathered} (d)/(dx)(y^2-2xy)=(d)/(dx)(16) \\ 2y*(dy)/(dx)-2(d)/(dx)(xy)=0 \\ 2y*(dy)/(dx)-2(x*(dy)/(dx)+1*y)=0 \\ 2y*(dy)/(dx)-2x*(dy)/(dx)-2y=0 \\ 2(y-x)*(dy)/(dx)-2y=0 \\ (y-x)*(dy)/(dx)-y=0 \\ (y-x)*(dy)/(dx)=y \\ (dy)/(dx)=(y)/(y-x) \end{gathered}`

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:

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

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.