Question

The slope of the tangent line to a curve at any point (x, y) on the curve is x/y. What is the equation of the curve if (4, 1) is a point on the curve?

x2 − y2 = 15
x2 + y2 = 15
x + y = 15 <- my answer
xy = 15

Answer 1

to get an x and y in the slope, because that was the result of a derivative, you must have had an equation
x^2=y^2 + k or
x^2-y^2=k
proof
2x-2y dy/dx=d/dx K=0
dy/dx=x/y

Answer 2

to get an x and y in the slope, because that was the result of a derivative, you must have had an equation
x^2=y^2 + k or
x^2-y^2=k
proof
2x-2y dy/dx=d/dx K=0
dy/dx=x/y
opps, you want the k for the funcion.
4^2-1^2=k
k=15
x^2-y^2=15