Find the slope of the tangent to the curve y^2=((x^2)/(xy-28)) at (8,4)

Question

the problem is stated as above. I'm just not sure where to start. 

Answer 1

y² = x²/(xy-28)

 

xy³-28y²= x²

 

y³+ 3xy²(dy/dx) - 56y(dy/dx) = 2x

 

When x = 8 and y = 4, we have 64 + 384(dy/dx) - 224(dy/dx) = 16

 

                                                                          160(dy/dx) = -48

 

                                                                                 dy/dx = -3/10 

Answer 2

Curve: y² = x²/(xy-28)

 

Let's get the slope field.

 

y²(xy - 28) = x²

 

xy³ - 28y² = x²

 

y³ + 3xy²y' -56yy' = 2x

 

(-3xy + 56)yy' = y³ - 2x

 

y' = (y³ - 2x)/[y(-3xy + 56)]     ← This is the slope field.

 

The slope at at (8,4) is:

 

(4³ - 2·8)/[4(-3·8·4 + 56)] = -3/10