Convert the equation to the standard form for a hyperbola by completing the square.
-$$4 y ^ { 2 } - 25 x ^ { 2 } + 8 y - 50 x - 121 = 0$$
A) $\frac { ( y - 1 ) ^ { 2 } } { 25 } - \frac { ( x - 1 ) ^ { 2 } } { 4 } = 1$
B) $\frac { ( y + 1 ) ^ { 2 } } { 25 } - \frac { ( x + 1 ) ^ { 2 } } { 4 } = 1$
C) $\frac { ( y + 1 ) ^ { 2 } } { 4 } - \frac { ( x + 1 ) ^ { 2 } } { 25 } = 1$
D) $\frac { ( \mathrm { x } - 1 ) ^ { 2 } } { 4 } - \frac { ( \mathrm { y } - 1 ) ^ { 2 } } { 25 } = 1$

Question

Quizplus · Accepted Answer

The Answer of Convert the equation to the standard form for a hyperbola by completing the square.
-$$4 y ^ { 2 } - 25 x ^ { 2 } + 8 y - 50 x - 121 = 0$$
A) $\frac { ( y - 1 ) ^ { 2 } } { 25 } - \frac { ( x - 1 ) ^ { 2 } } { 4 } = 1$
B) $\frac { ( y + 1 ) ^ { 2 } } { 25 } - \frac { ( x + 1 ) ^ { 2 } } { 4 } = 1$
C) $\frac { ( y + 1 ) ^ { 2 } } { 4 } - \frac { ( x + 1 ) ^ { 2 } } { 25 } = 1$
D) $\frac { ( \mathrm { x } - 1 ) ^ { 2 } } { 4 } - \frac { ( \mathrm { y } - 1 ) ^ { 2 } } { 25 } = 1$

Convert the Equation to the Standard Form for a Hyperbola 4y2−25x2+8y−50x−121=04 y ^ { 2 } - 25 x ^ { 2 } + 8 y - 50 x - 121 = 04y2−25x2+8y−50x−121=0

Convert the Equation to the Standard Form for a Hyperbola $4 y ^ { 2 } - 25 x ^ { 2 } + 8 y - 50 x - 121 = 0$