In the family x^2-α xy+y^2+4x+6y-β=0 , determine α and β so…

In the family $x^2-\alpha xy+y^2+4x+6y-\beta=0$, determine $\alpha$ and $\beta$ so that the curves pass through $A(2,0)$ and are parabolas. Sketch the parabolas.

Step 1: Conditions

Parabola:

\[ \delta=1-\tfrac{\alpha^2}{4}=0\ \Longrightarrow\ \alpha=\pm2 . \]

Through $A(2,0)$: $4+8-\beta=0$, so $\beta=12$.

Non-degenerate:

\[ \Delta=\begin{vmatrix}1&-\tfrac\alpha2&2\\-\tfrac\alpha2&1&3\\2&3&-12\end{vmatrix}=3\alpha^2-6\alpha-25 . \]

This gives $\Delta=-25$ for $\alpha=2$ and $\Delta=-1$ for $\alpha=-2$, both non-zero.

\[\alpha=\pm2,\quad\beta=12\]

Step 2: $\mathcal P_1$ ($\alpha=2$): $(x-y)^2+4x+6y-12=0$

Use the rotation $x=\tfrac{u-v}{\sqrt2}$, $y=\tfrac{u+v}{\sqrt2}$, with $u$ along $(1,1)$. Then $x-y=-\sqrt2\,v$ and $4x+6y=5\sqrt2\,u+\sqrt2\,v$, so

\[ 2v^2+\sqrt2\,v+5\sqrt2\,u-12=0\ \Longrightarrow\ 2\Big(v+\tfrac{\sqrt2}4\Big)^2=-5\sqrt2\Big(u-\tfrac{49\sqrt2}{40}\Big). \]

\[v'^2=-\tfrac{5\sqrt2}{2}\,u',\quad p=\tfrac{5\sqrt2}4,\quad V_1\left(\tfrac{59}{40},\tfrac{39}{40}\right),\quad\text{axis }x-y=\tfrac12\]

The parabola opens in the direction $-(1,1)$.

Step 3: $\mathcal P_2$ ($\alpha=-2$): $(x+y)^2+4x+6y-12=0$

With the same rotation, $x+y=\sqrt2\,u$, so

\[ 2u^2+5\sqrt2\,u+\sqrt2\,v-12=0\ \Longrightarrow\ 2\Big(u+\tfrac{5\sqrt2}{4}\Big)^2=-\sqrt2\Big(v-\tfrac{73\sqrt2}{8}\Big). \]

\[u'^2=-\tfrac{1}{\sqrt2}\,v',\quad p=\tfrac{\sqrt2}4,\quad V_2\left(-\tfrac{83}8,\tfrac{63}8\right),\quad\text{axis }x+y=-\tfrac52\]

The parabola opens in the direction $(1,-1)$.

Checks:

• $V_1$: $\tfrac14+\tfrac{236}{40}+\tfrac{234}{40}-12=0$;
• $V_2$: $\tfrac{25}4-\tfrac{83}2+\tfrac{189}4-12=0$;
• $A(2,0)$ lies on both curves.