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Let's find the derivative of that hyperbola in order to find a slope formula to help us with this equation.  We already have an x and y value.  The derivative is found this way:  [tex]y'= \frac{x(0)-3(1)}{x^2} [/tex] so  [tex]y'=- \frac{3}{x^2} [/tex].  The derivative supplies us with the slope formula we need to write the equation.  Sub in the x value of 3 to find what the slope is: [tex]y'=- \frac{3}{3^2}=- \frac{3}{9}=- \frac{1}{3} [/tex].  So in our slope-intercept equation, x = 3, y = 1, and m = -1/3.  Use these values to solve for b.  [tex]1=- \frac{1}{3}(3)+b [/tex]  so b = 2.  The equation, then, for the line tangent to that hyperbola at that given point is [tex]y=- \frac{1}{3}x+2 [/tex]
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The equation of the line tangent to the hyperbola at the point [tex](3,1)[/tex] is [tex]y = -\frac{1}{3}\cdot x + 2[/tex].

Procedure - Determination of a line tangent to a hyperbola

By analytical geometry we know that tangent lines are described by the line equation:

[tex]y = m\cdot x + b[/tex] (1)

Where:

  • [tex]x[/tex] - Independent variable.
  • [tex]m[/tex] - Slope.
  • [tex]b[/tex] - Intercept.
  • [tex]y[/tex] - Dependent variable.

By calculus we know that the slope of the tangent line is defined by the first derivative of the hyperbola evaluated at point of tangence:

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

If we know that [tex]x = 3[/tex] and [tex]y = 1[/tex], then the equation of the tangent line is:

Slope

[tex]m = -\frac{1}{3}[/tex]

Intercept

[tex]b = 1-\left(-\frac{1}{3} \right)\cdot (3)[/tex]

[tex]b = 2[/tex]

The equation of the line tangent to the hyperbola at the point [tex](3,1)[/tex] is [tex]y = -\frac{1}{3}\cdot x + 2[/tex]. [tex]\blacksquare[/tex]

To learn more on tangent lines, we kindly invite to check this verified question: https://brainly.com/question/8846313

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