Answer:
[tex]\dfrac{(y+2)^2}{4} -\dfrac{(x-4)^2}{9}=1[/tex]
Step-by-step explanation:
Given equation:
[tex]-4x^2 + 9y^2 + 32x + 36y - 64 = 0[/tex]
Collect like terms:
[tex]\implies 9y^2 + 36y -4x^2+ 32x - 64 = 0[/tex]
Move the constant to the right side of the equation:
[tex]\implies 9y^2 + 36y -4x^2+ 32x =64[/tex]
Factor out 9 from the terms in y, and 4 from the terms in x:
[tex]\implies 9(y^2 + 4y) -4(x^2-8x)=64[/tex]
Add the square of half the coefficient of the term in y and x inside the parentheses on the LHS, and the distributed values on the RHS:
[tex]\implies 9\left(y^2 + 4y+\left(\dfrac{4}{2}\right)^2 \right) -4\left(x^2-8x+\left(\dfrac{-8}{2}\right)^2 \right)=64+9\left(\dfrac{4}{2}\right)^2 \right+4\left(\dfrac{-8}{2}\right)^2 \right[/tex]
Simplify:
[tex]\implies 9(y^2 + 4y+4) -4(x^2-8x+16)=64+36-64[/tex]
[tex]\implies 9(y^2 + 4y+4) -4(x^2-8x+16)=36[/tex]
Factor the perfect trinomials:
[tex]\implies 9(y+2)^2 -4(x-4)^2=36[/tex]
Divide both sides by 36:
[tex]\implies \dfrac{9(y+2)^2}{36} -\dfrac{4(x-4)^2}{36}=\dfrac{36}{36}[/tex]
Simplify:
[tex]\implies \dfrac{\diagup\!\!\!\!9(y+2)^2}{\diagup\!\!\!\! 9\cdot 4} -\dfrac{\diagup\!\!\!\!4(x-4)^2}{9 \cdot \diagup\!\!\!\!4}=1[/tex]
[tex]\implies \dfrac{(y+2)^2}{4} -\dfrac{(x-4)^2}{9}=1[/tex]
Therefore, the function is a hyperbola with:
- Center: (4, -2)
- Vertices: (4, -4) and (4, 0)
- Co-vertices: (1, -2) and (7, -2)
- Foci: (4, -2±√13)
- Transverse axis: x = 4
- Conjugate axis: y = -2
- [tex]\textsf{Asymptotes}: \quad y =-2 \pm \left(\dfrac{2}{3}\right)(x-4)[/tex]
