Identify the conic by writing the equation in standard form. $4 x ^ { 2 } + 4 y ^ { 2 } + 8 x - 16 y - 12 = 0$
A) $\frac { ( x + 1 ) ^ { 2 } } { \frac { 2 } { 4 } } + \frac { ( y - 2 ) ^ { 2 } } { \frac { 2 } { 4 } } = 8$ ; ellipse
B) $( x + 1 ) ^ { 2 } + ( y - 2 ) ^ { 2 } = 4$ ; circle
C) $( x + 1 ) ^ { 2 } + ( y - 2 ) ^ { 2 } = 8$ ; circle
D) $( x + 1 ) ^ { 2 } + ( y - 4 ) ^ { 2 } = 8$ ; circle
E) $\frac { ( x + 1 ) ^ { 2 } } { \frac { 17 } { 4 } } + \frac { ( y - 2 ) ^ { 2 } } { \frac { 17 } { 4 } } = 8$ ; ellipse

Question

Identify the conic by writing the equation in standard form.  $4 x ^ { 2 } + 4 y ^ { 2 } + 8 x - 16 y - 12 = 0$ 
A) $\frac { ( x + 1 ) ^ { 2 } } { \frac { 2 } { 4 } } + \frac { ( y - 2 ) ^ { 2 } } { \frac { 2 } { 4 } } = 8$ ; ellipse
B) $( x + 1 ) ^ { 2 } + ( y - 2 ) ^ { 2 } = 4$ ; circle
C) $( x + 1 ) ^ { 2 } + ( y - 2 ) ^ { 2 } = 8$ ; circle
D) $( x + 1 ) ^ { 2 } + ( y - 4 ) ^ { 2 } = 8$ ; circle
E) $\frac { ( x + 1 ) ^ { 2 } } { \frac { 17 } { 4 } } + \frac { ( y - 2 ) ^ { 2 } } { \frac { 17 } { 4 } } = 8$ ; ellipse

Quizplus · Accepted Answer

The given equation can be rewritten as $(x+1)^2 + (y-2)^2 = 8$, which is the standard form of a circle with center $(-1, 2)$ and radius $\sqrt{8}$.

Identify the Conic by Writing the Equation in Standard Form 4x2+4y2+8x−16y−12=04 x ^ { 2 } + 4 y ^ { 2 } + 8 x - 16 y - 12 = 04x2+4y2+8x−16y−12=0

Identify the Conic by Writing the Equation in Standard Form $4 x ^ { 2 } + 4 y ^ { 2 } + 8 x - 16 y - 12 = 0$