Convert the equation to the standard form for a hyperbola by completing the square on x and y.
-$$4 x ^ { 2 } - 16 y ^ { 2 } - 16 x - 32 y - 64 = 0$$
A) $\frac { ( x - 2 ) ^ { 2 } } { 16 } - \frac { ( y + 1 ) ^ { 2 } } { 4 } = 1$
B) $\frac { ( x + 2 ) ^ { 2 } } { 16 } - \frac { ( y + 1 ) ^ { 2 } } { 4 } = 1$
C) $\frac { ( x - 2 ) ^ { 2 } } { 16 } - \frac { ( y - 1 ) ^ { 2 } } { 4 } = 1$
D) $\frac { ( x - 2 ) ^ { 2 } } { 4 } - \frac { ( y + 1 ) ^ { 2 } } { 16 } = 1$

Question

Quizplus · Accepted Answer

To convert the given equation to the standard form of a hyperbola by completing the square, we start with the given equation:$$4x^2 - 16y^2 - 16x - 32y - 64 = 0$$Group the x terms and the y terms:$$4x^2 - 16x - 16y^2 - 32y = 64$$Factor out the coefficients of the squared terms:$$4(x^2 - 4x) - 16(y^2 + 2y) = 64$$To complete the square, add and subtract the square of half the coefficient of x and y inside the parentheses (for x: $4/2 = 2$, $2^2 = 4$; for y: $2/2 = 1$, $1^2 = 1$):$$4[(x^2 - 4x + 4) - 4] - 16[(y^2 + 2y + 1) - 1] = 64$$Simplify and rearrange:$$4(x - 2)^2 - 16(y + 1)^2 = 64 + 16 - 16$$$$4(x - 2)^2 - 16(y + 1)^2 = 64$$Divide everything by 64 to get 1 on the right side:$$\frac{(x - 2)^2}{16} - \frac{(y + 1)^2}{4} = 1$$This matches option A.

Convert the Equation to the Standard Form for a Hyperbola 4x2−16y2−16x−32y−64=04 x ^ { 2 } - 16 y ^ { 2 } - 16 x - 32 y - 64 = 04x2−16y2−16x−32y−64=0

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