The slope of the tangent line to a curve at any point (x, y) on the curve is x divided by y. What is the equation of the curve if (2, 1) is a point on the curve?

a) x + y = 3
b) x^2 - y^2 = 3
c) x^2 + y^2 = 3
d) xy = 3

Respuesta :

The curve must have a positive slope in the first quadrant. The only choice that does is
  b) x^2 - y^2 = 3
Ver imagen sqdancefan

Answer:

Option b - [tex]x^2-y^2=3[/tex]

Step-by-step explanation:

Given : The slope of the tangent line to a curve at any point (x, y) on the curve is x divided by y.

To find : What is the equation of the curve if (2, 1) is a point on the curve?

Solution :

We know that,

The slope of the tangent line is the derivative.

So, [tex]\frac{dy}{dx}=\frac{x}{y}[/tex]

or [tex]yy'=x[/tex]

Subtract yy' on both side of the equation,

[tex]x-yy'=0[/tex]

Now, Integrating both side of the equation but keeping concept of implicit differentiation,

[tex](\frac{1}{2})x^2-(\frac{1}{2})y^2=c[/tex]

Where, c is the constant.

Solving the equation,

[tex](\frac{1}{2})(x^2-y^2)=c[/tex]

Substitute x=2 and y=1 to find c,

[tex](\frac{1}{2})(2^2-1^2)=c[/tex]

[tex](\frac{1}{2})(3)=c[/tex]

[tex]c=\frac{3}{2}[/tex]

Substitute c back in the equation,

[tex](\frac{1}{2})(x^2-y^2)=\frac{3}{2}[/tex]

Solving,

[tex]x^2-y^2=3[/tex]

Therefore, The required equation is [tex]x^2-y^2=3[/tex]

So, Option b is correct.