Why Intercepts and Symmetry Make Useful Checks
A sketch of a surface can look plausible while still placing the surface incorrectly or giving it the wrong balance. Coordinate-axis intercepts provide definite points the surface must pass through. Reflection symmetries tell us when reflecting a point across a particular plane must leave the surface unchanged.
These features are checks, not a complete description of a surface. A few intercepts and symmetries may be consistent with several different surfaces. But once an equation has suggested a shape, the checks can reveal a missed sign, a misplaced center, or an incorrect sketch.
Coordinate-Axis Intercepts
The \(x\)-axis consists of points \((x,0,0)\), the \(y\)-axis of points \((0,y,0)\), and the \(z\)-axis of points \((0,0,z)\). To find where a surface meets an axis, substitute the other two coordinates as zero into the surface equation and solve the resulting equation.
An axis restriction may have no real solution, one real solution, or more than one real solution. No real solution means the surface misses that axis; one solution means it touches the axis at one point; two solutions give two intercepts. If the restricted equation is true for every point on an axis, the whole axis lies in the surface, so its intersection is not a finite list of intercepts.
Keep track of which variable remains free. Solving for \(x\) with \(y=z=0\) gives points \((x,0,0)\), not points with \(x\) set to zero.
Worked Example: Intercepts of a Shifted Ellipsoid
Worked Example: Intercepts of a Shifted Ellipsoid
Find the coordinate-axis intercepts of \(\frac{(x-1)^2}{4}+\frac{y^2}{9}+\frac{z^2}{16}=1\), and check whether the origin lies on the surface.
On the \(x\)-axis, set \(y=0\) and \(z=0\): \[ \frac{(x-1)^2}{4}=1. \] Multiplying by \(4\) gives \((x-1)^2=4\), so \(x-1=\pm2\). Thus \(x=3\) or \(x=-1\), and the intercepts are \((3,0,0)\) and \((-1,0,0)\).
On the \(y\)-axis, set \(x=0\) and \(z=0\): \[ \frac{1}{4}+\frac{y^2}{9}=1. \] Therefore \(\frac{y^2}{9}=\frac{3}{4}\), so \(y^2=\frac{27}{4}\) and \(y=\pm\frac{3\sqrt{3}}{2}\). The intercepts are \((0,\frac{3\sqrt{3}}{2},0)\) and \((0,-\frac{3\sqrt{3}}{2},0)\).
On the \(z\)-axis, set \(x=0\) and \(y=0\): \[ \frac{1}{4}+\frac{z^2}{16}=1. \] Then \(\frac{z^2}{16}=\frac{3}{4}\), so \(z^2=12\) and \(z=\pm2\sqrt{3}\). The intercepts are \((0,0,2\sqrt{3})\) and \((0,0,-2\sqrt{3})\).
Finally, at the origin the left side of the original equation is \(\frac{1}{4}\), not \(1\). The origin is not on the surface. This is consistent with the intercept calculations: none of the three axes has the origin as its only intersection.
Reflection Symmetry from the Equation
Reflecting a point across the coordinate plane \(x=0\) changes \(x\) to \(-x\) while leaving \(y\) and \(z\) unchanged. Reflections across \(y=0\) and \(z=0\) similarly change the sign of \(y\) and \(z\), respectively. A surface is symmetric about a plane if reflecting any point of the surface across that plane produces another point on the surface.
If replacing \(x\) by \(-x\) leaves the equation unchanged, then the surface is symmetric about \(x=0\). More generally, if the replacement turns the left side into a nonzero constant multiple of the original left side, the equation still has the same solutions: multiplying both sides of an equation equal to zero by a nonzero constant does not change where it is zero. These are straightforward sufficient checks. When the equation does not pass such a check, do not assume symmetry; test whether reflection actually preserves the surface.
For an equation written in terms such as \((x-h)^2\), the relevant reflection plane may be shifted from \(x=0\). Reflection across \(x=h\) sends \(x\) to \(2h-x\). For instance, \(x-h\) becomes \(h-x=-(x-h)\), so its square is unchanged. An equation depending on \(x\) only through \((x-h)^2\) is therefore symmetric about \(x=h\), provided the other variables are unchanged by that reflection.
Central symmetry about the origin is a different test: reflect \((x,y,z)\) to \((-x,-y,-z)\) and check whether surface membership is preserved. It is not the same as symmetry about any one coordinate plane.
Worked Example: Coordinate-Plane Symmetry
Worked Example: Coordinate-Plane Symmetry
Test the symmetries of \(\frac{x^2}{4}+\frac{y^2}{9}-\frac{z^2}{16}=1\) about the three coordinate planes.
Reflect across \(x=0\): replace \(x\) by \(-x\). The term \(\frac{(-x)^2}{4}\) equals \(\frac{x^2}{4}\), so the equation is unchanged.
Reflect across \(y=0\): replace \(y\) by \(-y\). Since \(\frac{(-y)^2}{9}=\frac{y^2}{9}\), the equation is unchanged.
Reflect across \(z=0\): replace \(z\) by \(-z\). Since \(-\frac{(-z)^2}{16}=-\frac{z^2}{16}\), this equation is unchanged too. The surface is symmetric about all three coordinate planes.
Its intercepts also give useful checks. On the \(x\)-axis, \(y=z=0\) gives \(x^2/4=1\), hence \(x=\pm2\). On the \(y\)-axis, \(x=z=0\) gives \(y^2/9=1\), hence \(y=\pm3\). On the \(z\)-axis, \(x=y=0\) gives \(-z^2/16=1\), which has no real solution. Thus the surface crosses the \(x\)- and \(y\)-axes at equal positive and negative values and misses the \(z\)-axis.
Substitution verifies each listed crossing: \(x=\pm2,y=z=0\) makes the left side \(1\); \(y=\pm3,x=z=0\) also makes it \(1\). The absence of real \(z\)-intercepts follows because \(-z^2/16\leq0\), so it cannot equal \(1\).
A Short Justification: Why the Substitution Test Works
A reflection changes the coordinates of each point according to its geometric definition. To test whether a surface is preserved, ask whether the reflected coordinates satisfy its equation whenever the original coordinates do. If the equation is unchanged by the substitution, the answer is immediate.
Suppose \(F(x,y,z)=0\).
For reflection across \(x=0\), the point becomes \((-x,y,z)\).
If \(F(-x,y,z)=F(x,y,z)\), then \(F(-x,y,z)=0\), so the reflected point is also on the surface.
Reflecting twice returns the original point. Thus the same argument works in reverse, so membership is preserved in both directions.
This proves the symmetry test when the substitution leaves \(F\) unchanged. If it instead changes \(F\) by a nonzero constant factor, the zeros are also unchanged. The important geometric condition is preservation of the set of points satisfying the equation, not merely a visual resemblance between expressions.
Worked Example: A Shifted Reflection Plane
Worked Example: A Shifted Reflection Plane
Consider the surface \[ z=\frac{(x-2)^2}{4}+\frac{y^2}{9}-1. \] Find its coordinate-axis intercepts and identify its reflection symmetries among the planes \(x=0\), \(x=2\), \(y=0\), and \(z=0\).
On the \(x\)-axis, set \(y=z=0\): \[ 0=\frac{(x-2)^2}{4}-1. \] Thus \((x-2)^2=4\), giving \(x-2=\pm2\), so \(x=0\) or \(x=4\). The intercepts are \((0,0,0)\) and \((4,0,0)\).
On the \(y\)-axis, set \(x=z=0\): \[ 0=\frac{4}{4}+\frac{y^2}{9}-1=\frac{y^2}{9}. \] Therefore \(y=0\), giving the intercept \((0,0,0)\).
On the \(z\)-axis, set \(x=y=0\): \[ z=\frac{4}{4}+0-1=0. \] The intercept is again \((0,0,0)\).
For \(x=2\), replace \(x\) by \(4-x\). Then \(((4-x)-2)^2=(2-x)^2=(x-2)^2\), so the equation is unchanged: the surface is symmetric about \(x=2\). Replacing \(y\) by \(-y\) also leaves \(y^2\) unchanged, so it is symmetric about \(y=0\).
Replacing \(x\) by \(-x\) changes \((x-2)^2\) to \((x+2)^2\), so the equation does not pass the direct invariance check for \(x=0\). Replacing \(z\) by \(-z\) changes the left side while leaving the right side fixed; the equation does not pass the check for \(z=0\), either. The intercepts offer a quick consistency check on the shifted \(x\)-symmetry: the two \(x\)-intercepts, \(0\) and \(4\), are equally far from \(x=2\).
As a final verification, substituting \((4,0,0)\) gives a right side of \(\frac{(4-2)^2}{4}-1=1-1=0\), matching \(z=0\). Substituting \((0,0,0)\) gives \(\frac{(-2)^2}{4}-1=1-1=0\), also matching \(z=0\).
Use the Checks to Refine a Sketch
When interpreting a quadric equation, record the intercepts and symmetries before relying on a sketch. Then compare those facts with the drawing:
- Every reported axis intercept must lie on the surface.
- An axis with no real solution must not be drawn crossing the surface.
- For a symmetry plane, points on opposite sides must match under reflection.
- If symmetry is about a shifted plane such as \(x=h\), the balance is around that plane, not necessarily around \(x=0\).
These checks are especially valuable for shifted equations. A surface can be symmetric about a plane that does not pass through the origin. Treating every squared coordinate as evidence of symmetry about a coordinate plane can therefore misplace the sketch.
Common Mistake: Finding a Trace Instead of an Intercept
To find an \(x\)-axis intercept, both \(y\) and \(z\) must be set to zero. Setting only \(y=0\) leaves \(z\) free and restricts the surface to the plane \(y=0\); the resulting curve is a trace, not a list of \(x\)-axis intercepts.
For example, on \(\frac{x^2}{4}+\frac{y^2}{9}+\frac{z^2}{16}=1\), setting only \(y=0\) gives \(\frac{x^2}{4}+\frac{z^2}{16}=1\), an entire curve in the \(xz\)-plane. It would be wrong to call every point on that curve an \(x\)-axis intercept. To find those intercepts, also set \(z=0\); then \(\frac{x^2}{4}=1\), so \(x=\pm2\), and the intercepts are \((2,0,0)\) and \((-2,0,0)\).
Check Your Understanding
Use the definitions and checks above to answer these questions.
- For \(\frac{x^2}{9}+\frac{y^2}{4}-\frac{z^2}{16}=1\), find all coordinate-axis intercepts and identify any axis the surface does not meet.
- What coordinate substitution tests symmetry about the plane \(y=0\)?
- Why does dependence on \((x-3)^2\), rather than \(x^2\), suggest testing reflection about \(x=3\)?
- For \(z=(x-1)^2+y^2\), find the \(x\)-axis intercepts by setting both \(y\) and \(z\) to zero.
- Explain why setting only \(z=0\) does not, in general, find the \(x\)-axis intercepts.