Convert the equation to the standard form for an ellipse by completing the square on x and y.
-$$36 x ^ { 2 } + 16 y ^ { 2 } + 216 x - 64 y - 188 = 0$$
A) $\frac { ( x + 3 ) ^ { 2 } } { 16 } + \frac { ( y - 2 ) ^ { 2 } } { 36 } = 1$
B) $\frac { ( x - 2 ) ^ { 2 } } { 16 } + \frac { ( y + 3 ) ^ { 2 } } { 36 } = 1$
C) $\frac { ( x + 3 ) ^ { 2 } } { 36 } + \frac { ( y - 2 ) ^ { 2 } } { 16 } = 1$
D) $\frac { ( x - 3 ) ^ { 2 } } { 16 } + \frac { ( y + 2 ) ^ { 2 } } { 36 } = 1$

Question

Quizplus · Accepted Answer

To convert the given equation to the standard form of an ellipse, we complete the square for both $x$ and $y$. The given equation is $36x^2 + 16y^2 + 216x - 64y - 188 = 0$. For $x$, factor out 36: $36(x^2 + 6x) + 16y^2 - 64y - 188 = 0$. To complete the square, add and subtract $(6/2)^2 = 9$ inside the parenthesis, multiplied by 36: $36(x^2 + 6x + 9) + 16y^2 - 64y - 188 - 324 = 0$.For $y$, factor out 16: $36(x^2 + 6x + 9) + 16(y^2 - 4y) - 188 - 324 = 0$. To complete the square, add and subtract $(-4/2)^2 = 4$ inside the parenthesis, multiplied by 16: $36(x^2 + 6x + 9) + 16(y^2 - 4y + 4) - 188 - 324 - 64 = 0$.Simplifying, we get: $36(x + 3)^2 + 16(y - 2)^2 - 576 = 0$. Adding 576 to both sides and dividing by 576 gives: $\frac{(x + 3)^2}{16} + \frac{(y - 2)^2}{36} = 1$, which matches option A.

Convert the Equation to the Standard Form for an Ellipse 36x2+16y2+216x−64y−188=036 x ^ { 2 } + 16 y ^ { 2 } + 216 x - 64 y - 188 = 036x2+16y2+216x−64y−188=0

Convert the Equation to the Standard Form for an Ellipse $36 x ^ { 2 } + 16 y ^ { 2 } + 216 x - 64 y - 188 = 0$