Question

The slope of the tangent to a curve at any point (x, y) on the curve is -x/y. Find the equation of the curve if the point (3, −4) is on the curve.

Answer 1

`\bf \stackrel{\textit{slope or derivative}}{-\cfrac{x}{y}}\qquad (\stackrel{x}{3},\stackrel{y}{-4})\qquad \left. -\cfrac{x}{y} \right|_(3,-4)\implies -\cfrac{3}{-4}\implies \stackrel{\stackrel{slope}{m}}{\cfrac{3}{4}}\\\\ -------------------------------\\\\ \stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\implies y-(-4)=\cfrac{3}{4}(x-3) \implies y+4=\cfrac{3}{4}x-\cfrac{9}{4} \\\\\\ y=\cfrac{3}{4}x-\cfrac{9}{4}-4\implies y=\cfrac{3}{4}x-\cfrac{25}{4}`

Answer 2

The differential equation
dy/dx = -x/y
is a separable differential equation that has the solution
x^2 + y^2 = c

Substituting the given values for x and y, we find c=25.

The desired equation is
x^2 + y^2 = 25