Respuesta :
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.
Answer:
The correct option is D
Step-by-step explanation:
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