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Convert the Equation to the Standard Form for a Hyperbola 4x216y216x32y64=04 x ^ { 2 } - 16 y ^ { 2 } - 16 x - 32 y - 64 = 0

Question 4

Multiple Choice

Convert the equation to the standard form for a hyperbola by completing the square on x and y.
- 4x216y216x32y64=04 x ^ { 2 } - 16 y ^ { 2 } - 16 x - 32 y - 64 = 0


A) (x2) 216(y+1) 24=1\frac { ( x - 2 ) ^ { 2 } } { 16 } - \frac { ( y + 1 ) ^ { 2 } } { 4 } = 1
B) (x+2) 216(y+1) 24=1\frac { ( x + 2 ) ^ { 2 } } { 16 } - \frac { ( y + 1 ) ^ { 2 } } { 4 } = 1
C) (x2) 216(y1) 24=1\frac { ( x - 2 ) ^ { 2 } } { 16 } - \frac { ( y - 1 ) ^ { 2 } } { 4 } = 1
D) (x2) 24(y+1) 216=1\frac { ( x - 2 ) ^ { 2 } } { 4 } - \frac { ( y + 1 ) ^ { 2 } } { 16 } = 1

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