Find the centres of the curves: 1^° 3x^2-2xy+3y^2+4x+4y-4=0…

Find the centres of the curves: $1^\circ$ $3x^2-2xy+3y^2+4x+4y-4=0$; $2^\circ$ $2x^2-4xy+5y^2-8x+6=0$.

Step 1: Method

For $F=Ax^2+Bxy+Cy^2+Dx+Ey+F_0$ the centre solves

\[ 2Ax_0+By_0+D=0,\qquad Bx_0+2Cy_0+E=0 . \]

The solution is unique iff $4AC-B^2\ne0$. After translating to the centre, the constant becomes $F'=F(x_0,y_0)=\tfrac12(Dx_0+Ey_0)+F_0$.

Step 2: $1^\circ$

\[ 6x_0-2y_0+4=0,\quad -2x_0+6y_0+4=0\ \Longrightarrow\ x_0=y_0=-1 \]

Here $4AC-B^2=32$ and $F'=\tfrac12(-4-4)-4=-8$.

\[O_1(-1,-1),\qquad 3x'^2-2x'y'+3y'^2=8\]

The eigenvalues are $2$ and $4$, so rotating by $45^\circ$ gives $\tfrac{X^2}{4}+\tfrac{Y^2}{2}=1$: a real ellipse.

Step 3: $2^\circ$

\[ 4x_0-4y_0-8=0,\quad -4x_0+10y_0=0\ \Longrightarrow\ x_0=\tfrac{10}{3},\ y_0=\tfrac43 \]

Here $4AC-B^2=24$ and $F'=\tfrac12\left(-8\cdot\tfrac{10}{3}\right)+6=-\tfrac{22}{3}$.

\[O_1\left(\tfrac{10}{3},\tfrac43\right),\qquad 2x'^2-4x'y'+5y'^2=\tfrac{22}{3}\]

The eigenvalues are $1$, with direction $(2,1)$, and $6$, with direction $(1,-2)$. This gives $\tfrac{X^2}{22/3}+\tfrac{Y^2}{11/9}=1$: a real ellipse.