Write the equation of the hyperbola in standard form. Identify the center and vertices.
-$$4 y ^ { 2 } - 16 x ^ { 2 } + 40 y - 32 x + 20 = 0$$
A) $\frac { ( y - 5 ) ^ { 2 } } { 16 } - \frac { ( x - 1 ) ^ { 2 } } { 4 } = 1$
center: $( - 1 , - 5 )$; vertices: $( - 1 , - 9 ) , ( - 1 , - 1 )$
B) $\frac { ( y + 5 ) ^ { 2 } } { 16 } - \frac { ( x + 1 ) ^ { 2 } } { 4 } = 1$
center: $( - 1 , - 5 )$; vertices: $( - 1 , - 9 ) , ( - 1 , - 1 )$
C) $\frac { ( y + 5 ) ^ { 2 } } { 16 } - \frac { ( x + 1 ) ^ { 2 } } { 4 } = 1$
center: $( 1,5 )$; vertices: $( 1,1 ) , ( 1,9 )$
D) $\frac { ( y - 5 ) ^ { 2 } } { 16 } - \frac { ( x - 1 ) ^ { 2 } } { 4 } = 1$
center: $( 1,5 )$; vertices: $( 1,1 ) , ( 1,9 )$

Question

Quizplus · Accepted Answer

The given equation can be rewritten in standard form by completing the square for both $x$ and $y$. The correct standard form is $\frac { ( y + 5 ) ^ { 2 } } { 16 } - \frac { ( x + 1 ) ^ { 2 } } { 4 } = 1$, with the center at $(-1, -5)$ and vertices at $(-1, -9)$ and $(-1, -1)$.

Write the Equation of the Hyperbola in Standard Form 4y2−16x2+40y−32x+20=04 y ^ { 2 } - 16 x ^ { 2 } + 40 y - 32 x + 20 = 04y2−16x2+40y−32x+20=0

Write the Equation of the Hyperbola in Standard Form $4 y ^ { 2 } - 16 x ^ { 2 } + 40 y - 32 x + 20 = 0$