Question

Given \(f(x,y) = x^2 -2xy+ y^2\).
The complete contour of the equation \(f(x,y) = 1\) is described by the option(s) ___.

• A. A line, \(x-y= 1\)
• B. A circle with radius \(1\)
• C. An ellipse with length of major axis equal to \(1\)
• D. A line, \(x-y= -1\)

Hint:

Notice \(x^2-2xy+y^2=(x-y)^2\), so \(f=1\) means \((x-y)^2=1\), which splits into two lines, not a circle or ellipse.

Answer 1

The Correct Option is A, D

Step 1: Try to simplify f(x,y) before doing anything else.
The expression is
\[ f(x,y) = x^2-2xy+y^2 \]
This has exactly the pattern of a perfect square, \(a^2-2ab+b^2=(a-b)^2\), with \(a=x\) and \(b=y\). So
\[ f(x,y) = (x-y)^2 \]
Step 2: Write the contour equation using this simplified form.
The contour \(f(x,y)=1\) becomes
\[ (x-y)^2 = 1 \]
Step 3: Take the square root of both sides carefully.
Whenever we take a square root of an equation, we must keep both the positive and the negative root, since squaring either one gives back \(1\):
\[ x-y = \pm1 \]
This splits into two separate, independent linear equations,
\[ x-y=1 \qquad \text{and} \qquad x-y=-1 \]
Step 4: Recognize what these two equations represent geometrically.
Each equation \(x-y=c\) is a straight line of slope \(1\). Since the two lines have the same slope but different intercepts, they are two distinct parallel lines, one shifted from the other. Together, these two lines make up the complete contour \(f(x,y)=1\); neither line alone is the whole picture, but each one is a genuine part of it.
Step 5: Rule out the circle and the ellipse.
For the level set of a quadratic \(Ax^2+Bxy+Cy^2=1\) to be a circle or an ellipse, the quadratic form has to be positive definite in a way that does not collapse along any direction, which happens when the discriminant \(B^2-4AC\) is negative. Here \(A=1,\,B=-2,\,C=1\), so
\[ B^2-4AC = (-2)^2-4(1)(1) = 4-4 = 0 \]
A zero discriminant is the borderline, degenerate case: instead of a closed curve like a circle or ellipse, the quadratic factors exactly into a perfect square, which is precisely what produced the pair of parallel lines above. So options (B) and (C) describe the wrong family of curve altogether; \((x-y)^2=1\) can never trace out a circle or an ellipse, no matter how the radius or axis length is chosen.
Step 6: Confirm both lines are needed for the complete contour.
The question asks for the option(s) describing the complete contour. Since \((x-y)^2=1\) is satisfied exactly when \(x-y=1\) or \(x-y=-1\), both option (A) and option (D) are correct, and together they form the entire contour.
Final Answer:
\[ \boxed{x-y=1 \ \text{and}\ x-y=-1} \]