Diophantine Equation or Ellipse

Question

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 ?

Answer 1

Hint: One useful trick is completing the square. We have $2x^2-4x=2(x-1)^2-2$ and $3y^2-12y=3(y-2)^2-12$.

Remark: "Trick" is not really a suitable name for such a widely useful idea, which comes up in many branches of mathematics.

Answer 2

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 $y-2=0$.