By translation and rotation of the rectangular system xOy…
By translation and rotation of the rectangular system $xOy$, reduce $2x^2-xy+2y^2-x+\lambda=0$ to canonical form and write the transformation formulas.
Explanation of Method
A second-degree equation with parameter λ reduces to an ellipse, a point, or no real curve.
Step 1: Quadratic part
\[ A_0=\begin{pmatrix}2&-\tfrac12\\-\tfrac12&2\end{pmatrix},\qquad (2-\mu)^2=\tfrac14\ \Longrightarrow\ \mu_1=\tfrac32,\ \mu_2=\tfrac52. \]
The eigenvectors are $(1,1)$ for $\mu_1$ and $(-1,1)$ for $\mu_2$. Rotating by $\varphi=45^\circ$ gives
\[ x=\frac{u-v}{\sqrt2},\qquad y=\frac{u+v}{\sqrt2}. \]
Substituting: $\tfrac32u^2+\tfrac52v^2-\frac{u-v}{\sqrt2}+\lambda=0$.
Step 2: Translation
\[ \tfrac32\Big(u-\tfrac{\sqrt2}{6}\Big)^2-\tfrac1{12}+\tfrac52\Big(v+\tfrac{\sqrt2}{10}\Big)^2-\tfrac1{20}+\lambda=0. \]
With $u'=u-\tfrac{\sqrt2}6$, $v'=v+\tfrac{\sqrt2}{10}$ and $\tfrac1{12}+\tfrac1{20}=\tfrac2{15}$: $\tfrac32u'^2+\tfrac52v'^2=\tfrac2{15}-\lambda$.
Check with the centre equations $4x-y-1=0,\ -x+4y=0$: the centre is $\left(\tfrac4{15},\tfrac1{15}\right)$, and $F'=\lambda+\tfrac12\cdot(-1)\cdot\tfrac4{15}=\lambda-\tfrac2{15}$. $\checkmark$
Step 3: Canonical form and discussion
For $\lambda\lt \tfrac2{15}$: \[\frac{u'^2}{a^2}+\frac{v'^2}{b^2}=1,\qquad a^2=\frac{4-30\lambda}{45},\quad b^2=\frac{4-30\lambda}{75}\]
This is an ellipse with $a\gt b$ and $e=\sqrt{1-\tfrac{45}{75}}=\sqrt{\tfrac25}$ (independent of $\lambda$). For $\lambda=\tfrac2{15}$ the equation gives the single point $\left(\tfrac4{15},\tfrac1{15}\right)$. For $\lambda\gt \tfrac2{15}$ there are no real points (an imaginary ellipse).
Step 4: Transformation formulas
\[x=\frac{u'-v'}{\sqrt2}+\frac4{15},\qquad y=\frac{u'+v'}{\sqrt2}+\frac1{15}\]
- $\lambda=0$
- $\lambda=-\tfrac12$
- $\lambda=-1$
Result
\[\boxed{\,\tfrac32u'^2+\tfrac52v'^2=\tfrac2{15}-\lambda\,}\]
\[\boxed{\,\frac{u'^2}{a^2}+\frac{v'^2}{b^2}=1,\qquad a^2=\frac{4-30\lambda}{45},\quad b^2=\frac{4-30\lambda}{75}\,}\]
\[\boxed{\,x=\frac{u'-v'}{\sqrt2}+\frac4{15},\qquad y=\frac{u'+v'}{\sqrt2}+\frac1{15}\,}\]
Remark
Correction: the “i.e.” form $a^2=\tfrac43\left(\tfrac1{10}-\tfrac\lambda2\right)$ is wrong. It equals $\tfrac2{15}-\tfrac{2\lambda}3$, whereas $a^2=\tfrac23\left(\tfrac2{15}-\lambda\right)=\tfrac{4}{45}-\tfrac{2\lambda}{3}$. Correct: $a^2=\tfrac{4-30\lambda}{45}$, $b^2=\tfrac{4-30\lambda}{75}$.
Note: the source uses $x=\tfrac{u+v}{\sqrt2},\ y=\tfrac{u-v}{\sqrt2}$, whose matrix has determinant $-1$. That is a reflection, not a rotation (the same issue as in Problem 160). It gives the same canonical equation, but it changes orientation. The formulas above are a genuine rotation by $45^\circ$. The centre $\left(\tfrac4{15},\tfrac1{15}\right)$ is the same in both.