An ellipse with length of major axis equal to \(1\)
A line, \(x-y= -1\)
Show Solution
The Correct Option isA, D
Solution and Explanation
Step 1: Write $f$ as a quadratic form and look at its matrix.
$f(x,y)=x^2-2xy+y^2$ can be written as $\mathbf{v}^TM\mathbf{v}$ with $\mathbf{v}=\begin{pmatrix}x\\y\end{pmatrix}$ and
\[
M = \begin{pmatrix}1 & -1\\ -1 & 1\end{pmatrix}
\] Step 2: Find the eigenvalues of M.
The characteristic equation is $\det(M-\lambda I)=0$:
\[
(1-\lambda)^2-1=0 \ \Rightarrow\ \lambda^2-2\lambda=0 \ \Rightarrow\ \lambda(\lambda-2)=0
\]
so $\lambda=0$ or $\lambda=2$. Step 3: Read off what a zero eigenvalue means.
An ellipse or circle needs both eigenvalues of its quadratic form to be strictly positive (a positive definite form), so the level curve is bounded in every direction. Here one eigenvalue is exactly $0$, which means the quadratic form does not grow at all along the eigenvector direction for $\lambda=0$ (this direction turns out to be along $x=y$). The curve is therefore unbounded along that direction, ruling out a circle (option B) and an ellipse (option C) immediately, without needing to factor anything. Step 4: Change variables to see the shape directly.
Let $u=x-y$ (the $\lambda=2$ direction) and $v=x+y$ (the $\lambda=0$ direction). Then $f=x^2-2xy+y^2=(x-y)^2=u^2$, which does not involve $v$ at all. The equation $f=1$ becomes $u^2=1$, that is $u=\pm1$: two straight lines parallel to the $v$-axis (the free direction), at $u=1$ and $u=-1$. Step 5: Translate back to x and y.
$u=x-y$, so $u=1$ gives $x-y=1$ and $u=-1$ gives $x-y=-1$. Both are needed together to form the full contour, since the original equation $u^2=1$ is satisfied by either value of $u$.
Options (A) and (D) both correctly describe the two lines making up the contour; options (B) and (C) are wrong because the quadratic form is degenerate (rank $1$, one zero eigenvalue), not positive definite.
\[
\boxed{x-y=1\ \text{and}\ x-y=-1}
\]
