MATH SOLVE

4 months ago

Q:
# The slope of the tangent to a curve at any point (x, y) on the curve is x divided by y. Find the equation of the curve if the point (2, −3) is on the curve. x2 + y2 = 13 x2 + y2 = 25 x2 − y2 = −5 x2 − y2 = 5

Accepted Solution

A:

Answer:[tex]x^2-y^2=-5[/tex]Step-by-step explanation:The problem tells us that the slope of the tangent to a curve at any point [tex](x, y)[/tex] on the curve is x divided by y, that is:[tex]m=\frac{x}{y}[/tex]We also know that the point [tex](2,-3)[/tex] is on the curve. By taking a look on the options this point lies on both equations, namely:[tex]x^2+y^2=13 \ because \ (2)^2+(-3)^2=13 \\ \\ x^2-y^2=-5 \ because \ (2)^2-(-3)^2=-5[/tex]We know that the derivative is the slope of the tangent line to the graph of the function at a given point. So taking the derivative of both equations we have:[tex]\frac{d}{dx}(x^2+y^2)=\frac{d}{dx}(13) \\ \\ \therefore 2x+2y\frac{dy}{dx}=0 \\ \\ \therefore m=\frac{dy}{dx}=-\frac{x}{y}[/tex]And:[tex]\frac{d}{dx}(x^2-y^2)=\frac{d}{dx}(-5) \\ \\ \therefore 2x-2y\frac{dy}{dx}=0 \\ \\ \therefore m=\frac{dy}{dx}=\frac{x}{y}[/tex]So [tex]x^2-y^2=-5[/tex] also meets the requirement of the condition the slope of the tangent to a curve at any point (x, y) on the curve is x divided by y. Therefore this is the correct option.