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

Question

Quizplus · Accepted Answer

To convert the given equation to the standard form of a hyperbola by completing the square, we first group the x's and y's together and move the constant to the other side: $y^2 - 4y - 4x^2 - 16x = 16$. Then, we complete the square for both x and y terms. For the y terms: $y^2 - 4y + 4$ (adding 4 inside the parenthesis and to the other side to balance), and for the x terms: $-4(x^2 + 4x + 4)$ (adding 16 inside the parenthesis but since it's multiplied by -4, we add -64 to the other side to balance). So, the equation becomes $(y^2 - 4y + 4) - 4(x^2 + 4x + 4) = 16 + 4 - 64$, which simplifies to $(y - 2)^2 - 4(x + 2)^2 = -44$. Dividing everything by -44 gives us the standard form of a hyperbola: $\frac{(y - 2)^2}{4} - \frac{(x + 2)^2}{4} = 1$, which matches option A.

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

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