Take (x′+2) for x and (y′+2) for y to change the origin. So, new equation can be written as, 3(x′+2)2+2(x′+2)(y′+2)+3(y′+2)2−18(x′+2)−22(y′+2)+50]=0 Solve and obtain, 3x′2+2x′y′+3y′2−1=0 By dropping the suffices the equation can be written as, 3x2+2xy+3y2−1=0 For to turn the axes through an angle of
π
3
, in the counter clockwise sense, we should write
x′−√3y′
2
for x and
√3x′+y′
2
for y. So, [3(
x′−√3y′
2
)2+2(
x′−√3y′
2
)(
√3x′+y′
2
)+3(
√3x′+y′
2
)2−1]=0 By dropping the suffices and solving the equation, [3(x2+3y2−2√3xy)+2(√3x2−2xy−√3y2)+3(3x2+y2+2√3xy)−4]=0 (12+2√3)x2−4xy+(12−2√3)y2−4=0 (6+√3)x2−2xy+(6−√3)y2−2=0