Write Equations of Rotated Conics in Standard Form
-$$4 x ^ { 2 } - 4 x y + y ^ { 2 } - 6 x + 12 = 0$$
A) $\left( y ^ { \prime } + \frac { 6 \sqrt { 5 } } { 25 } \right) ^ { 2 } = \frac { 6 \sqrt { 5 } } { 25 } x ^ { \prime } - \frac { 264 } { 125 }$
B) $y ^ { \prime 2 } = - 6 x ^ { \prime }$
C) $\frac { x ^ { \prime 2 } } { 6 } + y ^ { \prime 2 } = 1$
D) $\frac { x ^ { \prime 2 } } { 5 } + \frac { y ^ { \prime 2 } } { 6 } = 1$ )

Question

Quizplus · Accepted Answer

The given equation $4 x ^ { 2 } - 4 x y + y ^ { 2 } - 6 x + 12 = 0$ represents a rotated conic. To find its equation in standard form, we need to eliminate the xy-term, which suggests a rotation of the coordinate system. The correct transformation leads to option A, which is the standard form of the equation after rotation and completing the square.

Write Equations of Rotated Conics in Standard Form - 4x2−4xy+y2−6x+12=04 x ^ { 2 } - 4 x y + y ^ { 2 } - 6 x + 12 = 04x2−4xy+y2−6x+12=0

Write Equations of Rotated Conics in Standard Form
- $4 x ^ { 2 } - 4 x y + y ^ { 2 } - 6 x + 12 = 0$