Answer:
[tex]\dfrac{(x-1)^2}{9}-\dfrac{(y-2)^2}{4}=1[/tex]
Step-by-step explanation:
Given equation:
[tex]4x^2-9y^2-8x+36y-68=0[/tex]
This is an equation for a horizontal hyperbola.
To complete the square for a hyperbola
Arrange the equation so all the terms with variables are on the left side and the constant is on the right side.
[tex]\implies 4x^2-8x-9y^2+36y=68[/tex]
Factor out the coefficient of the x² term and the y² term.
[tex]\implies 4(x^2-2x)-9(y^2-4y)=68[/tex]
Add the square of half the coefficient of x and y inside the parentheses of the left side, and add the distributed values to the right side:
[tex]\implies 4\left(x^2-2x+\left(\dfrac{-2}{2}\right)^2\right)-9\left(y^2-4y+\left(\dfrac{-4}{2}\right)^2\right)=68+4\left(\dfrac{-2}{2}\right)^2-9\left(\dfrac{-4}{2}\right)^2[/tex]
[tex]\implies 4\left(x^2-2x+1\right)-9\left(y^2-4y+4\right)=36[/tex]
Factor the two perfect trinomials on the left side:
[tex]\implies 4(x-1)^2-9(y-2)^2=36[/tex]
Divide both sides by the number of the right side so the right side equals 1:
[tex]\implies \dfrac{4(x-1)^2}{36}-\dfrac{9(y-2)^2}{36}=\dfrac{36}{36}[/tex]
Simplify:
[tex]\implies \dfrac{(x-1)^2}{9}-\dfrac{(y-2)^2}{4}=1[/tex]
Therefore, this is the standard equation for a horizontal hyperbola with:
- center = (1, 2)
- vertices = (-2, 2) and (4, 2)
- co-vertices = (1, 0) and (1, 4)
- [tex]\textsf{Asymptotes}: \quad y = -\dfrac{2}{3}x+\dfrac{8}{3} \textsf{ and }y=\dfrac{2}{3}x+\dfrac{4}{3}[/tex]
- [tex]\textsf{Foci}: \quad (1-\sqrt{13}, 2) \textsf{ and }(1+\sqrt{13}, 2)[/tex]
