Transform 3x^2-10xy+3y^2+32=0 by rotating the axes through…

Transform $3x^2-10xy+3y^2+32=0$ by rotating the axes through $45^\circ$ and identify the curve.

Explanation of Method

A 45° rotation reduces the given equation to a canonical hyperbola.

Step 1: Substitution

\[ x=\frac{X-Y}{\sqrt2},\qquad y=\frac{X+Y}{\sqrt2}\quad\Longrightarrow\quad x^2+y^2=X^2+Y^2,\quad xy=\frac{X^2-Y^2}{2}. \]

Write $3x^2+3y^2-10xy=3(x^2+y^2)-10xy$: $3(X^2+Y^2)-5(X^2-Y^2)+32=-2X^2+8Y^2+32=0$.

\[\frac{X^2}{16}-\frac{Y^2}{4}=1\]

Step 2: Elements

A hyperbola with $a=4$, $b=2$, $c=2\sqrt5$, $e=\tfrac{\sqrt5}{2}$. Its real axis is the line $y=x$. The vertices $(X,Y)=(\pm4,0)$ are the old points $\pm(2\sqrt2,2\sqrt2)$.

The asymptotes $Y=\pm\tfrac12X$ come from the quadratic part: \[3x^2-10xy+3y^2=(3x-y)(x-3y)=0\ \Longrightarrow\ y=3x,\quad y=\tfrac13x.\]

$X$ $Y$ $y=3x$ $y=\tfrac x3$ $(2\sqrt2,2\sqrt2)$ $(-2\sqrt2,-2\sqrt2)$

After rotating by $45^\circ$: $\dfrac{X^2}{16}-\dfrac{Y^2}{4}=1$. Real axis on $y=x$; asymptotes $y=3x$, $y=x/3$ (orange).

Result

\[\boxed{\,\frac{X^2}{16}-\frac{Y^2}{4}=1\,}\]