Calculus BC NEED HELP ASAP PLS!!

Question

The slope of the tangent to a curve at any point (x, y) on the curve is  . Find the equation of the curve if the point (2, -3) is on the curve. (5 points)

Answer Choices

	x2 + y2 = 13
	x2 + y2 = 25
	x2 - y2 = -5
	x2 - y2 = 5

Answer 1

Draw a circle centered at the origin. While you can solve using implicit differentiation that dy/dx = -x/y, note that you can easily reason out the sign. In the first and third quadrants, the slope of the tangent is negative (obvious from the picture),and x/y is positive. In the second and fourth quadrants, the slope of the tangent is positive and x/y is negative. Thus, we don't have a circle, so we are left with the two hyperbolas. Plug in x and y and you get that the solution is x^2 - y^2 = -5

Now, the fancier way using implicit differentiation.

x^2 + y^2 = c (either 13 or 25)

2xdx + 2ydy = 0

2ydy = -2xdx

dy/dx = -2x/2y = -x/y

For the hyperbolas,

x^2 - y^2 = c

2xdx - 2ydy = 0

2ydy = 2xdx

dy/dx = 2x/2y = x/y