Find the center and vertices of the hyperbola. $12 x ^ { 2 } - 4 y ^ { 2 } - 72 x - 16 y + 44 = 0$
A)center: (3, -2), vertices: (3, -4), (3, 0)
B)center: (-3, 2), vertices: (-3, 0), (-3, 4)
C)center: (3, -2), vertices: (1, -2), (5,-2)
D)center: (2, -3), vertices: (0, -3), (4, -3)
E)center: (-3,2), vertices: (-5, 2), (-1, 2)

Question

Quizplus · Accepted Answer

First, complete the square for both $x$ and $y$ terms in the equation $12x^2 - 4y^2 - 72x - 16y + 44 = 0$. Group $x$ and $y$ terms: $(12x^2 - 72x) - (4y^2 + 16y) = -44$. Divide by the coefficients of $x^2$ and $y^2$ to prepare for completing the square: $x^2 - 6x$ and $-y^2 - 4y$. Complete the square by adding and subtracting $(6/2)^2 = 9$ inside the $x$ group and $(-4/2)^2 = 4$ inside the $y$ group, considering the coefficients outside the parentheses: $12(x^2 - 6x + 9) - 4(y^2 + 4y + 4) = -44 + 12(9) - 4(4)$. Simplify to get the standard form of the hyperbola equation: $12(x - 3)^2 - 4(y + 2)^2 = 44 + 108 - 16$, which simplifies to $12(x - 3)^2 - 4(y + 2)^2 = 136$. Divide through by 136 to get $ \frac{(x - 3)^2}{\frac{136}{12}} - \frac{(y + 2)^2}{\frac{136}{4}} = 1$. The center of the hyperbola is at $(3, -2)$, and the vertices are found by adding and subtracting the square root of $\frac{136}{12}$ (the $a^2$ term under the $x$ part) from the $x$-coordinate of the center, resulting in vertices at $(3 \pm \sqrt{\frac{136}{12}}, -2)$ or simplified to $(3 \pm 2, -2)$, which are $(1, -2)$ and $(5, -2)$.

Find the Center and Vertices of the Hyperbola 12x2−4y2−72x−16y+44=012 x ^ { 2 } - 4 y ^ { 2 } - 72 x - 16 y + 44 = 012x2−4y2−72x−16y+44=0

Find the Center and Vertices of the Hyperbola $12 x ^ { 2 } - 4 y ^ { 2 } - 72 x - 16 y + 44 = 0$