Answer:
[tex]d.\:ellipse;\:3x^2+y^2+6x-6y+3=0[/tex].
Step-by-step explanation:
The given conic has equation;
[tex]3x^2+y^2=9[/tex]
Divide through by 9.
[tex]\Rightarrow \frac{3x^2}{9}+\frac{y^2}{9}=\frac{9}{9}[/tex]
[tex]\Rightarrow \frac{x^2}{3}+\frac{y^2}{9}=1[/tex].
This is an ellipse centered at the origin;
This ellipse has been translated so that the center is now at;
[tex](-1,3)[/tex]
The translated ellipse has equation;
[tex]\frac{(x+1)^2}{3}+\frac{(y-3)^2}{9}=1[/tex].
Clear the fraction;
[tex]3(x+1)^2+(y-3)^2=9[/tex].
Expand;
[tex]3(x^2+2x+1)+(y^2-6y+9)=9[/tex].
[tex]3x^2+6x+3+y^2-6y+9=9[/tex].
Write in general form;
[tex]3x^2+y^2+6x-6y+9-9+3=0[/tex].
[tex]3x^2+y^2+6x-6y+3=0[/tex].
The correct choice is D
