Equations Can Describe Curves Without Giving a Single Formula for y
A familiar way to describe a curve is as the graph of a function, such as \(y=f(x)\). But many curves are more naturally described by an equation involving both coordinates. The equation \(x^2+y^2=1\), for example, describes the unit circle without selecting one value of \(y\) for each \(x\).
To understand such an equation, read it as a condition on points. Every ordered pair \((x,y)\) that satisfies the condition belongs to the curve; every pair that does not satisfy it lies outside the curve. This viewpoint is useful even when solving the equation for \(y\) would produce multiple formulas or would require restricting the possible \(x\)-values.
In particular, an equation \(F(x,y)=0\) defines the zero level set of \(F\). The phrase implicit curve is often used for a curve described this way, rather than by an explicit formula \(y=f(x)\). The equation identifies the entire set of points; it does not promise that the set is the graph of one function.
Relations, Graphs, and Branches
A relation between \(x\) and \(y\) can be viewed as a set of ordered pairs. The relation defined by \(x^2+y^2=1\), for instance, contains precisely those pairs whose coordinates satisfy that equation. By contrast, the graph of a function \(f\) on a domain \(D\) is the set of pairs \((x,f(x))\) with \(x\in D\). For a graph to represent a function, each allowed \(x\)-value must have exactly one corresponding \(y\)-value.
This gives a practical test. A relation is the graph of a function of \(x\) only if no vertical line meets it in more than one point. If a vertical line meets the relation twice, then the relation as a whole cannot be the graph of one function \(y=f(x)\). It may still be made up of separate parts, called branches, each of which is a function graph on an appropriate domain.
Solving an equation for \(y\) can help reveal its branches, but it is not required to define the curve. When solving does produce two formulas, both may be needed to represent the full relation.
Worked Example: The Unit Circle and Its Two Branches
Worked Example: The Unit Circle and Its Two Branches
Interpret \(x^2+y^2=1\) as a level set, and find its branches as graphs of \(x\).
Define \(F(x,y)=x^2+y^2\). The given equation is \(F(x,y)=1\), so its points are exactly the pairs whose squared coordinates sum to \(1\). Geometrically, \(x^2+y^2\) is the square of the distance from \((x,y)\) to the origin. Thus the equation says that the distance is \(1\): the set is the circle of radius \(1\) centered at the origin.
To identify the graph branches, solve for \(y\):
A real value of \(y\) requires \(1-x^2\geq 0\), which is equivalent to \(-1\leq x\leq 1\). Taking both square roots gives the two branches:
For example, let \(x=\frac35\). Then
Both resulting points satisfy the original equation:
At most \(x\)-values strictly between \(-1\) and \(1\), the circle therefore contains two points with that \(x\)-coordinate: one on the upper branch and one on the lower branch. At \(x=1\) and \(x=-1\), the two formulas give the same point at that endpoint. The full circle is one relation, not one function graph.
Why the Circle Equation Describes a Geometric Locus
The connection between the equation \(x^2+y^2=1\) and the circle follows from the distance formula. The distance from \((x,y)\) to the origin is \(\sqrt{x^2+y^2}\). It equals \(1\) exactly when
Because both sides are nonnegative, squaring this equation gives \(x^2+y^2=1\), and if \(x^2+y^2=1\), taking the nonnegative square root gives \(\sqrt{x^2+y^2}=1\). Thus the distance condition and the level-set equation describe exactly the same points. The equation is not merely an instruction to calculate \(y\); it is a precise description of a geometric locus.
Worked Example: The Relation \(xy=1\)
Worked Example: The Relation \(xy=1\)
Describe the relation \(xy=1\), determine its branches, and verify two points on it.
Define \(F(x,y)=xy\). The equation \(F(x,y)=1\) selects all points whose coordinate product is \(1\). If \(x=0\), then \(xy=0\) for every \(y\), so no point on this relation has \(x=0\). For \(x\neq 0\), solving for \(y\) gives
The domain of this graph is \(x\neq 0\). It splits into two branches: one for \(x>0\), where \(y>0\), and one for \(x<0\), where \(y<0\). For instance, setting \(x=2\) gives \(y=\frac12\), and setting \(x=-2\) gives \(y=-\frac12\). Check each point in the original equation:
Each branch is a function graph on its own domain, but the full relation has points on both sides of the \(y\)-axis. It cannot be described by a single formula with a domain that includes \(x=0\), because the relation contains no point with \(x=0\). In this example, solving for \(y\) produces one formula, but its domain separates the relation into two branches.
Worked Example: A Hyperbola with Two Choices of y
Worked Example: A Hyperbola with Two Choices of y
Describe the set defined by \(x^2-y^2=1\), find where its graph branches can exist, and verify points on both sides of the plane.
The equation is the level set \(F(x,y)=1\) for \(F(x,y)=x^2-y^2\). Rearranging to isolate the square of \(y\) gives
For real \(y\), the right side must be nonnegative. Therefore \(x^2-1\geq 0\), so \(x^2\geq 1\), or \(x\leq -1\) or \(x\geq 1\). On those domains, the two formulas are
Take \(x=2\). Then \(y^2=2^2-1=3\), giving \(y=\sqrt3\) or \(y=-\sqrt3\). Substitution confirms both points:
Take \(x=-2\). Again \(y^2=(-2)^2-1=4-1=3\), so \(y=\sqrt3\) or \(y=-\sqrt3\). The checks are
These computations show that the relation includes points with positive and negative \(x\), and for each of these chosen \(x\)-values it includes two different \(y\)-values. The formulas describe its upper and lower branches, while the original equation describes the entire set at once.
How to Read an Implicit Equation
When an equation in \(x\) and \(y\) is presented as a curve, separate two questions:
List the points \((x,y)\) that satisfy it. This set-based interpretation works whether or not \(y\) can be isolated.
Check whether any \(x\)-value occurs with more than one \(y\)-value. If it does, the relation is not one function graph.
If useful, solve for \(y\), keep both signs when square roots appear, and state the domain on which each formula is real.
These questions should not be confused. An implicit equation defines its relation before any rearrangement takes place. Rearranging may reveal a graph representation, but it must preserve all the points and include the necessary domain restrictions.
Common Pitfall: Keeping Only One Branch
A frequent error is to take a square root and keep only its positive value. For the unit circle, the incorrect calculation is
This formula describes only the upper branch. For example, at \(x=0\) it gives \(y=1\), but the original relation also contains \((0,-1)\), since
The correction is to retain both square roots:
The domain restriction matters as well: if \(|x|>1\), then \(1-x^2<0\), so there is no real \(y\) on the circle for that \(x\). Whenever isolating \(y\) uses a square, remember that a squared equation can have positive and negative solutions, and check both against the original relation.
Why Level-Set Descriptions Matter
Level sets give a compact way to describe a whole geometric object, including all its branches. The circle equation \(x^2+y^2=1\) does not favor the upper half over the lower half, and \(xy=1\) includes both its positive and negative branches. This makes level-set notation useful whenever a curve is naturally specified by a condition on both coordinates.
It also keeps the distinction clear between an object and one way of representing part of it. A relation is the complete set of qualifying points. A branch is a portion that can be written as the graph of a function on a specified domain. An equation can define the relation even when no single-valued formula \(y=f(x)\) represents it all.
Check Your Understanding
Use the equation as a condition on ordered pairs before deciding whether it is one function graph.
- For \(x^2+y^2=4\), what is the level-set function and value? What are the possible \(y\)-values when \(x=0\)?
- Why does \(x^2+y^2=1\) fail the vertical line test at \(x=0\)?
- For \(xy=1\), explain why \(x=0\) is excluded and identify the signs of \(x\) and \(y\) on each branch.
- For \(x^2-y^2=1\), what restriction on \(x\) is required for real values of \(y\)?
- When solving an equation for \(y\) produces a square root, why might one formula fail to describe the full relation?
