Solving Level-Curve Equations

From a Height to a Planar Equation

A function $z=f(x,y)$ assigns a height to each point $(x,y)$ in its domain. If we specify one height $c$, the points of the graph at that height must satisfy $f(x,y)=c$. This equation involves only $x$ and $y$, so it describes a set in the plane.

Solving that planar equation is the central task here. The algebra tells us whether the level set is empty, a single point, a circle, a hyperbola, or another simple set. Its features then guide a sketch. In particular, sums and differences of squares behave differently, so the sign and value of $c$ matter.

Definition: For a function $f$ and a real number $c$, the level curve at height $c$ is the set $$ \{(x,y): f(x,y)=c\}. $$ It consists of all points in the domain of $f$ where the function has value $c$. Depending on $f$ and $c$, this set need not actually be a curve: it may be empty or consist of a single point.

Why Substituting the Height Works

The graph of $z=f(x,y)$ consists of the points $(x,y,z)$ satisfying $z=f(x,y)$. A horizontal plane at height $c$ has equation $z=c$. A point belongs to both the graph and that plane exactly when $z=f(x,y)$ and $z=c$, which together mean $f(x,y)=c$. Thus solving $f(x,y)=c$ gives precisely the $(x,y)$-coordinates of the horizontal slice. This explains why a three-dimensional slicing question becomes a planar equation.

z=f(x,y),\qquad z=c \quad\Longrightarrow\quad f(x,y)=c.

For example, if $f(x,y)=x^2+y^2$ and the chosen height is $c=9$, substitution gives $x^2+y^2=9$. The original function determines the equation; the specified height determines its right-hand side.

A Reliable Solving-and-Sketching Routine

For the two kinds of functions considered below, a careful solution can be organized into a few steps.

Substitute the height.
Write $f(x,y)=c$, using the specified value of $c$.

Rearrange and classify.
Use the sign and size of $c$ to determine what kind of set the equation describes, or whether it has any real solutions.

Find sketching features.
Use intercepts and, for hyperbolas, vertices and asymptotes to place and orient the branches.

Check the result.
Substitute a proposed point into the equation and verify that its left-hand side equals $c$.

Level Curves of a Sum of Squares

For $f(x,y)=x^2+y^2$, the level equation is $x^2+y^2=c$. Since squares are nonnegative, $x^2+y^2\geq 0$ for every real pair $(x,y)$. This immediately rules out solutions when $c<0$. When $c=0$, both squares must be zero, so the only solution is $(0,0)$. When $c>0$, the equation is a circle centered at the origin with radius $\sqrt{c}$.

Result: The equation $x^2+y^2=c$ has no real solutions if $c<0$; it gives the single point $(0,0)$ if $c=0$; and it gives a circle centered at the origin with radius $\sqrt{c}$ if $c>0$.

Worked Example: A Circle at Height 9

Let $f(x,y)=x^2+y^2$, and find and sketch the level curve at height $c=9$.

Step 1: Substitute the height.

$$x^2+y^2=9.$$

Step 2: Identify the curve. This has the form $x^2+y^2=r^2$, where $r^2=9$, so $r=3$. The curve is a circle centered at $(0,0)$ with radius $3$.

Step 3: Locate useful points. Setting $y=0$ gives $x^2=9$, hence $x=\pm3$. Setting $x=0$ gives $y^2=9$, hence $y=\pm3$. These four intercepts are $(3,0)$, $(-3,0)$, $(0,3)$, and $(0,-3)$. Sketch the circle through them.

Check: At $(3,0)$, $x^2+y^2=3^2+0^2=9$; at $(0,-3)$, $x^2+y^2=0^2+(-3)^2=9$. Both points lie on the stated level curve.

Level Curves of a Difference of Squares

For $f(x,y)=x^2-y^2$, substitution gives $x^2-y^2=c$. Unlike a sum of squares, this expression can be positive, negative, or zero: the value depends on which squared coordinate is larger.

If $c>0$, divide by $c$ to obtain $x^2/c-y^2/c=1$. This is a horizontal hyperbola: its vertices are $(\sqrt{c},0)$ and $(-\sqrt{c},0)$. If $c<0$, write $c=-a$ with $a>0$. Rearranging gives $y^2-x^2=a$, or $y^2/a-x^2/a=1$. This is a vertical hyperbola with vertices $(0,\sqrt{a})$ and $(0,-\sqrt{a})$. When $c=0$, factor the equation as $(x-y)(x+y)=0$; the level set is the pair of lines $y=x$ and $y=-x$, not a hyperbola.

Result: For $x^2-y^2=c$, a positive $c$ gives a horizontal hyperbola, a negative $c$ gives a vertical hyperbola, and $c=0$ gives the two intersecting lines $y=x$ and $y=-x$.

For nonzero $c$, the corresponding hyperbola has asymptotes $y=x$ and $y=-x$. These lines indicate the directions the branches approach far from the origin; they are not themselves points of the level curve. For instance, on either line $y=x$ or $y=-x$, one has $x^2-y^2=0$, so no point on either proposed asymptote lies on the level curve $x^2-y^2=4$.

Worked Example: A Horizontal Hyperbola at Height 4

Let $f(x,y)=x^2-y^2$, and find and sketch its level curve at height $c=4$.

Step 1: Substitute and put the equation in standard form.

x^2-y^2=4,\qquad \frac{x^2}{4}-\frac{y^2}{4}=1.

This is a horizontal hyperbola. Its vertices are $(2,0)$ and $(-2,0)$. Setting $x=0$ would require $-y^2=4$, which has no real solution, so there are no $y$-axis intercepts. The asymptote directions are $y=x$ and $y=-x$. Sketch two branches opening to the right and left from the vertices.

Check: At $(2,0)$, $x^2-y^2=2^2-0^2=4$. As another point on the curve, $(2\sqrt{2},2)$ gives $(2\sqrt{2})^2-2^2=8-4=4$. In contrast, every point on either proposed asymptote satisfies $x^2-y^2=0$, so none lies on this level curve.

Worked Example: A Vertical Hyperbola at Height Negative 9

Let $f(x,y)=x^2-y^2$, and solve and sketch the level curve at $c=-9$.

Step 1: Substitute the height and rearrange.

x^2-y^2=-9,\qquad y^2-x^2=9,\qquad \frac{y^2}{9}-\frac{x^2}{9}=1.

This is a vertical hyperbola. Its vertices are $(0,3)$ and $(0,-3)$. There are no $x$-axis intercepts: putting $y=0$ in the original equation gives $x^2=-9$, which has no real solution. Its asymptote directions are $y=x$ and $y=-x$. Sketch two branches opening upward and downward.

Check: At $(0,3)$, $x^2-y^2=0^2-3^2=-9$. Also, $(4,5)$ lies on the curve because $4^2-5^2=16-25=-9$.

Zero Height and a Sign Check

The level $c=0$ for the difference-of-squares function is worth solving explicitly. Factoring gives $x^2-y^2=(x-y)(x+y)=0$. A product is zero exactly when at least one factor is zero, so either $x-y=0$, which means $y=x$, or $x+y=0$, which means $y=-x$. The two lines intersect at the origin, and every point on either line has function value zero.

Worked Example: A Common Intercept Mistake

Suppose the equation $x^2-y^2=4$ is incorrectly said to have $y$-intercepts $(0,2)$ and $(0,-2)$. Check the claim by setting $x=0$:

0^2-y^2=4,\qquad -y^2=4,\qquad y^2=-4.

No real number has square $-4$, so these proposed intercepts do not exist. Direct substitution confirms the error: at $(0,2)$, $x^2-y^2=0-4=-4$, not $4$; at $(0,-2)$, the value is also $0-4=-4$. The correction is to use the vertices $(2,0)$ and $(-2,0)$, for which $x^2-y^2=4$. The branches open horizontally.

Why Solving Carefully Matters

Replacing the function value with a specified height is simple, but the resulting equation can change type with $c$. A negative value is impossible for $x^2+y^2$, while it produces a vertical hyperbola for $x^2-y^2$. At zero, the sum of squares gives a point, whereas the difference of squares gives two lines. Treating all these cases as the same kind of curve leads to incorrect sketches.

Use the original equation as a final test. A point on a level curve must make the function equal the specified height, not merely satisfy a guessed geometric description. In particular, proposed intercepts should be checked by setting the other coordinate to zero and solving over the real numbers. This small check catches many sign and orientation errors.

Check Your Understanding

For each question, solve the level equation over the real numbers and state the information that determines its sketch.

For $f(x,y)=x^2+y^2$, what is the level set at $c=0$, and what is it at $c=-1$?
For $f(x,y)=x^2+y^2$, identify the center and radius of the level curve at $c=16$.
For $f(x,y)=x^2-y^2$, rewrite the level equation at $c=9$ in standard form and give its vertices.
For $f(x,y)=x^2-y^2$, rewrite the level equation at $c=-4$ in standard form and state its orientation.
Factor the level equation $x^2-y^2=0$. Which two lines form its level set?

What You'll Learn