Find the centre of the curve 2x^2+3y^2-4x+6y-7=0 , write its…

Find the centre of the curve $2x^2+3y^2-4x+6y-7=0$, write its equation in the translated system, and identify the curve.

Step 1: Centre

\[ F_x=4x-4=0,\qquad F_y=6y+6=0\ \Longrightarrow\ O_1(1,-1). \]

Equivalently, completing the squares: $2(x-1)^2+3(y+1)^2=7+2+3=12$.

Step 2: New equation

With $x=x'+1$, $y=y'-1$ the linear terms cancel. The new constant is $F(1,-1)=2+3-4-6-7=-12$.

\[2x'^2+3y'^2=12\quad\Longleftrightarrow\quad\frac{x'^2}{6}+\frac{y'^2}{4}=1\]

Step 3: Elements

An ellipse with the following elements:

• semi-axes $a=\sqrt6$ (along $O_1x'$) and $b=2$;
• $c=\sqrt{a^2-b^2}=\sqrt2$ and $e=\tfrac{c}{a}=\tfrac{\sqrt3}{3}$;
• foci $(1\pm\sqrt2,\,-1)$ in the old coordinates;
• vertices $(1\pm\sqrt6,-1)$ and $(1,1)$, $(1,-3)$.