<s>[INST] Diophantine Equation or Ellipse I guess this is something to do with circle
The question is:
"Given x and y are real numbers, such that $2x^2 + 3y^2 - 4x - 12y = -14$, find xy."
What is the trick ?
 [/INST] Rewrite
$$2x^2-4x+3y^2-12y=-14$$
$$2(x^2-2x)+3(y^2-4y)=-14$$
$$2(x^2-2x+1-1)+3(y^2-4y+4-4)=-14$$
$$2(x^2-2x+1)-2+3(y^2-4y+4)-12=-14$$
$$2(x-1)^2+3(y-2)^2-14=-14$$
$$2(x-1)^2+3(y-2)^2=0$$
When is a sum of two squares equal to zero? Only if both squares are equal to zero. So one can conclude $x-1=0$ and $y-2=0$.
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