Convert the equation to the standard form for a hyperbola by completing the square on x and y.
-$$9 x ^ { 2 } - 4 y ^ { 2 } + 18 x - 16 y - 43 = 0$$
A) $\frac { ( x + 1 ) ^ { 2 } } { 4 } - \frac { ( y + 2 ) ^ { 2 } } { 9 } = 1$
B) $\frac { ( x - 1 ) ^ { 2 } } { 4 } - \frac { ( y + 2 ) ^ { 2 } } { 9 } = 1$
C) $\frac { ( x + 1 ) ^ { 2 } } { 4 } - \frac { ( y - 2 ) ^ { 2 } } { 9 } = 1$
D) $\frac { ( x + 1 ) ^ { 2 } } { 9 } - \frac { ( y + 2 ) ^ { 2 } } { 4 } = 1$

Question

Quizplus · Accepted Answer

To convert the given equation to the standard form of a hyperbola, complete the square for both x and y terms. For x: $9(x^2 + 2x)$ becomes $9((x+1)^2 - 1)$. For y: $-4(y^2 + 4y)$ becomes $-4((y+2)^2 - 4)$. Substitute back and simplify to get the standard form: $\frac{(x+1)^2}{4} - \frac{(y+2)^2}{9} = 1$.

Convert the Equation to the Standard Form for a Hyperbola 9x2−4y2+18x−16y−43=09 x ^ { 2 } - 4 y ^ { 2 } + 18 x - 16 y - 43 = 09x2−4y2+18x−16y−43=0

Convert the Equation to the Standard Form for a Hyperbola $9 x ^ { 2 } - 4 y ^ { 2 } + 18 x - 16 y - 43 = 0$