The slope of the tangent to a curve at any point (x, y) on the curve is negative 1 times x divided by y . Find the equation of the curve if the point (3, -4) is on the curve.

Question

A. x + y = -1

B. x^2 + y^2 = 5

C. x^2 + y^2 = 25

D. x^2 - y^2 = -7

Answer 1

dy/dx = -x/y

Write this as a differential equation

y dy = x dx

y² + C = x²

9 + C = 16

C=-7

x² + y² = -7

Answer 2

By condition we have differential equation: dy/dx = - x/y; By seporation variables we get: ydy = - xdx;

By integration we have: ∫ydy = - ∫ xdx; y²/2 = - x²/2 + C.

To get C we use conditions: y(3) = - 4. 16/2 = - 9/2 + C, C = 25/2.

So, y²/2 = - x²/2 + 25/2 or x² + y² = 25;

Answer is C.