Rotate the axes so that the equation 5x^2-4xy+2y^2-24=0 has…

Rotate the axes so that the equation $5x^2-4xy+2y^2-24=0$ has no $xy$ term. Identify the curve.

Explanation of Method

A rotation by arctan 2 removes the xy term and reveals an ellipse.

Step 1: Angle of rotation

For $Ax^2+Bxy+Cy^2+\dots$ (here $A=5,\ B=-4,\ C=2$) the $xy$ term disappears when

\[ \tan2\varphi=\frac{B}{A-C}=\frac{-4}{3}. \]

Choosing $\tan\varphi=2$ (then $\tan2\varphi=\tfrac{4}{1-4}=-\tfrac43$ ✓): $\cos\varphi=\tfrac{1}{\sqrt5}$, $\sin\varphi=\tfrac{2}{\sqrt5}$, $\varphi=\arctan2\approx63.43^\circ$.

Step 2: New equation

\[ x=\frac{X-2Y}{\sqrt5},\qquad y=\frac{2X+Y}{\sqrt5}. \]

\[ 5x^2=\tfrac{5X^2-20XY+20Y^2}{5},\quad -4xy=\tfrac{-8X^2+12XY+8Y^2}{5},\quad 2y^2=\tfrac{8X^2+8XY+2Y^2}{5}. \]

Summing: $\tfrac{5X^2+0\cdot XY+30Y^2}{5}=X^2+6Y^2$. Hence

\[X^2+6Y^2=24\quad\Longleftrightarrow\quad\frac{X^2}{24}+\frac{Y^2}{4}=1\]

An ellipse with $a=2\sqrt6$, $b=2$, $c=\sqrt{20}=2\sqrt5$, $e=\sqrt{\tfrac56}$. The coefficients $1$ and $6$ are the eigenvalues of $\begin{pmatrix}5&-2\\-2&2\end{pmatrix}$: $\lambda^2-7\lambda+6=0$.

$X$ $Y$ $\varphi$ $a=2\sqrt6$ $b=2$

Rotation by $\varphi=\arctan2\approx63.43^\circ$: the ellipse $\dfrac{X^2}{24}+\dfrac{Y^2}{4}=1$, major axis along the direction $(1,2)$.

Result

\[\boxed{\,X^2+6Y^2=24\quad\Longleftrightarrow\quad\frac{X^2}{24}+\frac{Y^2}{4}=1\,}\]

Remark

The other solution $\tan\varphi=-\tfrac12$ ($\varphi\approx-26.57^\circ$) is also correct. It gives $6X^2+Y^2=24$, the same ellipse with the roles of the axes swapped.