Find the equation of the tangent line to the follow curve at the indicated points. y^2=(x^2/(xy-4)) at (4,2)

Question

Find the equation of the tangent line to the follow curve at the indicated points.

y^2=(x^2/(xy-4)) at (4,2)

Back of my textbook states that the answer is y=2

Answer 1

The first thing that you need to do is find the slope of the curve at that specific point. My assumption is that this is a calculus questions so you understand how to implicitly differentiate that equation. It is as follows:

 

y² = x²/(xy-4)    

 

2ydy = (2xdx(xy-4) - x²(xdy+ydx))/ (xy-4)²           Use the quotient rule to differentiate the right side of the equation

 

2ydy = (2x²ydx - 8xdx - x³dy - x²ydx)/ (xy-4)2   Simplify

 

2ydy(xy-4)² + x³dy = (2x²y - 8x - x²y)dx

 

(2y(xy-4)² + x³)dy = (2x²y - 8x - x²y)dx

 

dy/dx = (2x²y - 8x - x²y) / (2y(xy-4)² + x³)

 

Now that we have taken the derivative, we plug in the given point in order to find the slope at that point.

 

dy/dx = (64 - 32 - 32) / (64 + 64)

 

dy/dx = 0 / 128

 

dy/dx = 0

 

We have figured out that our slope at the given point is 0. We can now use point slope form in order to figure out the equation. 

 

y - y₁ = m(x - x₁)

y - 2 = 0(x - 4)

y - 2 = 0

y = 2

 

This equation of the line makes sense. The derivative shows that the slope is zero which means we have found a critical point. A point on the graph where the slope changes from negative to positive or vice versa. Basically, we have found a local minimum or a local maximum. As a result, we would expect the equation of the line tangent to this point to be a horizontal line.