AP EAMCET Engineering 2018 Apr 22 Shift 2 Paper

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