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Question

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

Answer 1

dy/dx = x/y

ydy/dx = x

∫ydy = ∫xdx

y²/2 = x²/2 + C₁

y² = x² + C₂

At (3,1)

1 = 9 + C₂ --> C₂ = -8

y = ±√(x²-8)

Answer 2

dy/dx = x/y

ydy = xdx

1/2y² = 1/2x² + C

y² = x² + C₁ thru (3 , 1): 1 = 9 + C₁ , C₁ = -8

y² = x² - 8 , in explicit form y = ±√(x² - 8)

In std form for a hyperbola: x²/8 - y²/8 = 1