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

Question

Quizplus · Accepted Answer

To convert the given equation to the standard form of a hyperbola, we need to complete the square for both x and y terms. Grouping the x and y terms together and moving the constant to the other side gives us: $$9y^2 + 18y - 4x^2 + 16x = 43$$. Dividing the linear coefficients by 2 and squaring gives the numbers to add and subtract for completing the square: for y, $(\frac{18}{2*9})^2 = 1$, and for x, $(\frac{16}{2*(-4)})^2 = 4$. Adding and subtracting these inside their respective groups and rearranging gives: $$9(y^2 + 2y + 1) - 4(x^2 - 4x + 4) = 43 + 9 - 16$$, which simplifies to $$9(y + 1)^2 - 4(x - 2)^2 = 36$$. Dividing through by 36 to get 1 on the right side gives the standard form of the hyperbola: $$\frac{(y + 1)^2}{4} - \frac{(x - 2)^2}{9} = 1$$.

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

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