The slope of the tangent line to a curve at any point (x, y) on the curve is x divided by y. What is the equation of the curve if (3, 1) is a point on the curve?

Answer 1

slope = x/y = 3/1 = 3

so the equation of the tangent at the point (3,1) is
y-1 = 3(x-3)
y = 3x -8 or 3x - y - 8 = 0

Answer 2

y' = x/y
you can recognize that as an hyperbola

x^2-y^2 = a^2
since (3,1) is on the curve,
9-1 = a^2

x^2-y^2 = 8

Answer 3

Maybe I should become a politician.
Never actually answer the question, rather give an answer to some other question.
Thanks for the catch Steve, that is two for one evening, I should get more sleep.