How Separate Terms Shape a Contour
A function of two variables assigns a height to every point in the \(xy\)-plane. A contour records the points that have one selected height. When the function is a sum of a term depending only on \(x\) and a term depending only on \(y\), the level equation separates into those two contributions. Studying how the terms combine makes it easier to predict and sketch the contours.
The same basic form can produce very different curves. A sum of two positive square terms gives closed contours, while changing one term to a negative square can give hyperbolas. The signs and sizes of the terms matter: one should solve the level equation and check the resulting points rather than assume that every separable sum has the same contour shape.
A Reliable Sketching Strategy
For a chosen level \(c\), the contour equation is \(g(x)+h(y)=c\). Each point on the contour must contribute a value \(g(x)\) from the \(x\)-term and a value \(h(y)\) from the \(y\)-term whose sum is \(c\). Depending on the functions, it may be possible to solve for \(y\), for \(x\), or for one variable in terms of the other. But any such solution must respect the domain of the functions and the restrictions imposed by square roots or other operations.
For a practical sketch, first choose several representative contour levels. Then find intercepts by setting \(y=0\) or \(x=0\), respectively, and identify any symmetries directly from the equation. Check whether the equation has any solutions at all for the chosen level. Finally, use the equation and key points together to sketch the curve; intercepts alone do not determine its shape.
Write \(g(x)+h(y)=c\) for the selected contour level \(c\).
Use the ranges and signs of the terms to see whether solutions are possible.
Set one coordinate to zero to find intercepts, and test other convenient values if needed.
Use the solved equation, key points, and symmetries to identify the curve and its components.
Worked Example: A Sum of Two Squares
Worked Example: \(f(x,y)=x^2+y^2\)
Find and sketch the contour at height \(c=9\), and describe what happens at other levels.
The level equation is obtained by setting \(f(x,y)=9\):
To find the \(x\)-intercepts, set \(y=0\). Then \(x^2=9\), so \(x=3\) or \(x=-3\), giving \((3,0)\) and \((-3,0)\). To find the \(y\)-intercepts, set \(x=0\). Then \(y^2=9\), so \(y=3\) or \(y=-3\), giving \((0,3)\) and \((0,-3)\).
For example, at \(x= \frac{3}{\sqrt{2}}\), the equation requires \(y^2=9-\frac{9}{2}=\frac{9}{2}\), so \(y=\frac{3}{\sqrt{2}}\) or \(y=-\frac{3}{\sqrt{2}}\). Thus \((\frac{3}{\sqrt{2}},\frac{3}{\sqrt{2}})\) and \((\frac{3}{\sqrt{2}},-\frac{3}{\sqrt{2}})\) are also points on the contour. Substitution verifies the first point: \(\frac{9}{2}+\frac{9}{2}=9\); for the second, \(\frac{9}{2}+(-\frac{3}{\sqrt{2}})^2=\frac{9}{2}+\frac{9}{2}=9\).
The equation says that the squared distance from \((x,y)\) to the origin is \(9\), so the contour is a circle of radius \(3\). The equation is unchanged if either coordinate changes sign or if \(x\) and \(y\) are exchanged, which is consistent with the four intercepts and the circular shape.
For a general level, \(x^2+y^2=c\). If \(c>0\), this is a circle of radius \(\sqrt{c}\). If \(c=0\), both squares must be zero, so the only point is \((0,0)\). If \(c<0\), there are no solutions because a sum of squares cannot be negative. In particular, for \(c=0\), substitution gives \(0^2+0^2=0\), as required; for \(c<0\), no point can satisfy the equation.
Why Different Coefficients Make Ellipses
Consider a positive weighted sum of squares, \(x^2+4y^2\). Both terms are nonnegative, but the \(y\)-term grows four times as quickly as \(y^2\). On a fixed positive level, this means that less change in \(y\) is needed to use up the available level than change in \(x\). The resulting contour is narrower in the \(y\)-direction.
To justify the result, divide the level equation \(ax^2+by^2=c\) by \(c\), which is positive. This gives \(\frac{a}{c}x^2+\frac{b}{c}y^2=1\). Since \(a\) and \(b\) are positive, rewrite the coefficients as reciprocals of positive squared lengths: \(\frac{x^2}{c/a}+\frac{y^2}{c/b}=1\). Setting \(y=0\) gives \(ax^2=c\), hence \(x=\pm\sqrt{c/a}\); setting \(x=0\) gives \(by^2=c\), hence \(y=\pm\sqrt{c/b}\). These are the two axis lengths that determine the ellipse’s horizontal and vertical extent. If \(c=0\), the only solution is the origin; if \(c<0\), there are no solutions, because \(ax^2+by^2\geq 0\).
Worked Example: An Ellipse with Unequal Widths
Worked Example: \(f(x,y)=x^2+4y^2\)
Find and sketch the contour at height \(c=4\), then compare it with the contour at height \(c=16\).
At height \(4\), the equation is:
Setting \(y=0\) gives \(x^2=4\), so the \(x\)-intercepts are \((2,0)\) and \((-2,0)\). Setting \(x=0\) gives \(4y^2=4\), so \(y^2=1\), and the \(y\)-intercepts are \((0,1)\) and \((0,-1)\). Dividing the equation by \(4\) gives \(\frac{x^2}{4}+\frac{y^2}{1}=1\), identifying an ellipse that extends twice as far horizontally as vertically.
As an additional check, take \(x=1\). The level equation becomes \(1+4y^2=4\), so \(4y^2=3\) and \(y=\pm\frac{\sqrt{3}}{2}\). Substituting either value gives \(1+4(\frac{\sqrt{3}}{2})^2=1+4(\frac{3}{4})=4\). These points lie between the \(x\)-axis and the top and bottom intercepts, as expected.
At height \(16\), the equation is \(x^2+4y^2=16\). Its \(x\)-intercepts satisfy \(x^2=16\), so \(x=\pm4\); its \(y\)-intercepts satisfy \(4y^2=16\), so \(y=\pm2\). Dividing by \(16\) yields \(\frac{x^2}{16}+\frac{y^2}{4}=1\). The ellipse has grown in both directions, but it is still twice as wide horizontally as vertically. Checking \((0,2)\) gives \(0^2+4(2^2)=16\), confirming the upper intercept.
The coefficient \(4\) belongs to the \(y\)-term, so a given level allows a smaller magnitude of \(y\) than of \(x\). Specifically, at level \(4\), the maximum absolute \(x\)-value is \(2\), while the maximum absolute \(y\)-value is \(1\).
When a Separable Sum Gives a Hyperbola
Separability alone does not guarantee a circle or an ellipse. The sign of each term also matters. For example, \(x^2-y^2\) is still a sum of an \(x\)-term and a \(y\)-term, since it can be written \(g(x)+h(y)\) with \(g(x)=x^2\) and \(h(y)=-y^2\). But the negative contribution allows one squared term to balance the other, producing open curves rather than closed ones.
Worked Example: \(f(x,y)=x^2-y^2\)
Describe the contours at levels \(c=1\), \(c=-1\), and \(c=0\), paying careful attention to the components at level \(1\).
At height \(1\), the equation is \(x^2-y^2=1\). Solving for \(x\) gives \(x=\sqrt{1+y^2}\) or \(x=-\sqrt{1+y^2}\). Since \(1+y^2\geq1\), the first equation gives \(x\geq1\), and the second gives \(x\leq-1\). The contour therefore has a right branch, contained in \(x\geq1\), and a left branch, contained in \(x\leq-1\). These are its two connected components.
Equivalently, solving for \(y\) gives \(y=\sqrt{x^2-1}\) or \(y=-\sqrt{x^2-1}\), where \(|x|\geq1\). These equations describe the upper and lower portions, respectively. Each portion occurs on both the left and right branches; neither the upper portion nor the lower portion is a single connected branch of the hyperbola. The contour meets the \(x\)-axis at \((1,0)\) and \((-1,0)\), found by setting \(y=0\). It has no \(y\)-intercept, because setting \(x=0\) would require \(-y^2=1\), which has no real solution.
For a check, the point \((\sqrt{2},1)\) lies on the right branch because \(\sqrt{2}\geq1\). Substitution gives \((\sqrt{2})^2-1^2=2-1=1\). The point \((-\sqrt{2},1)\) lies on the left branch, and its substitution also gives \((-\sqrt{2})^2-1^2=2-1=1\). These checks show that the upper portion includes points on both connected branches.
At height \(-1\), the equation is \(x^2-y^2=-1\), or \(y^2-x^2=1\). Solving for \(y\) gives \(y=\sqrt{1+x^2}\) or \(y=-\sqrt{1+x^2}\). The first equation has \(y\geq1\), and the second has \(y\leq-1\), so in this case the connected components are the upper branch and the lower branch. The intercepts are \((0,1)\) and \((0,-1)\). For example, \((1,\sqrt{2})\) satisfies the equation because \(1^2-(\sqrt{2})^2=1-2=-1\).
At height \(0\), factor the equation \(x^2-y^2=0\) as \((x-y)(x+y)=0\). Thus the contour is the union of the two lines \(y=x\) and \(y=-x\), which meet at the origin. The origin checks correctly: \(0^2-0^2=0\).
Common Pitfall: Treating Every Sum as a Sum of Squares
A tempting but incorrect shortcut is to see two squared variables and assume the contour is an ellipse. For \(x^2-y^2=1\), one might incorrectly write \(x^2+y^2=1\) and sketch a circle. That changes the sign of the \(y^2\)-term and therefore changes the function. For example, the point \((0,1)\) satisfies \(x^2+y^2=1\), but for the actual level equation it gives \(0^2-1^2=-1\), not \(1\). The correction is to preserve every sign when setting \(f(x,y)=c\), then use that equation to find intercepts, restrictions, and branches.
It is also important not to confuse two different ways of describing a curve with its connected components. For \(x^2-y^2=1\), the expressions \(y=\sqrt{x^2-1}\) and \(y=-\sqrt{x^2-1}\) describe upper and lower portions. The two connected branches, however, are the left and right components, distinguished by \(x\leq-1\) and \(x\geq1\). Always identify components from the full equation, not just from the sign chosen when solving for one variable.
Why the Terms Matter
Each term controls how a coordinate contributes to the chosen height. When both terms are positive multiples of squares, as in \(ax^2+by^2\) with \(a,b>0\), increasing either coordinate’s magnitude increases the sum; positive levels give closed ellipses, with their widths determined by the coefficients. When one squared term is subtracted, increases in one coordinate can be balanced by increases in the other, and the contours can extend without closing. These conclusions rely on the actual signs and coefficients in the level equation.
For a general separable sum, the individual terms need not be squares. The equation \(g(x)+h(y)=c\) may lead to a different curve, and the method is still to select a level, solve or analyze the equation, and verify key points. The familiar circle, ellipse, and hyperbola shapes arise from particular choices of \(g\) and \(h\); they are useful models, not a rule that every separable sum has one of those shapes.
Check Your Understanding
Use the level equation in each question, and check any proposed point by substitution.
- For \(f(x,y)=x^2+y^2\), what is the contour at height \(c=4\), and what are its four intercepts?
- For \(f(x,y)=x^2+9y^2\), find the intercepts of the contour at height \(c=9\). Which coordinate direction has the smaller extent?
- For \(f(x,y)=x^2-y^2\), describe the connected components of the contour at height \(c=4\).
- For \(x^2-y^2=1\), explain why \(y=\sqrt{x^2-1}\) describes upper portions rather than one connected branch.
- Why does \(x^2+4y^2=c\) have no real solutions when \(c<0\)?