Why Signs Deserve a Second Look
A quadric can change substantially when a single sign changes. A positive squared term changed to negative may turn an ellipsoid into a hyperboloid, and a sign on the other side of a paraboloid equation determines which way it opens. A mistaken sign can also make an equation appear to describe a real surface when it has no real points.
The risk is greatest when rearranging an equation or completing squares. The goal here is to track the signs that determine the quadric’s type and orientation, then check the result against the original equation.
Read the Signs Relative to the Whole Equation
For a centered quadric with a nonzero constant, first isolate the squared terms and divide by the constant, taking care to divide every term by the same number. The familiar patterns then apply. Three positive squared terms equal to \(1\) describe an ellipsoid. Two positive terms and one negative term equal to \(1\) describe a hyperboloid of one sheet; the axis is the variable with the negative term. One positive term and two negative terms equal to \(1\) describe a hyperboloid of two sheets; its axis is the variable with the positive term.
These patterns only apply in the stated normalized forms. A constant of \(0\), missing squared terms, or other special coefficients may produce a different or degenerate surface. Those cases should not be forced into one of the patterns above.
One useful safeguard is to distinguish two operations that can look similar:
- Multiplying the entire equation by \(-1\) preserves the same surface. Every term, including the right-hand side, changes sign.
- Changing the sign of one squared term generally changes the surface. The relative signs, and therefore potentially the type or axis, are different.
Why the Sign Pattern Determines the Type
Each squared expression is nonnegative. That simple fact explains why an all-positive sum cannot equal a negative number, and why changing the sign of one term can make a previously impossible equation have real solutions.
For the mixed-sign form \(X^2/a^2+Y^2/b^2-Z^2/c^2=1\), with \(a,b,c>0\), one can solve for the negative term:
At \(Z=0\), the equation allows points satisfying \(X^2/a^2+Y^2/b^2=1\). For sufficiently large \(|Z|\), it also allows points, because the right side of the rearranged equation is then positive. The standard form is a hyperboloid of one sheet with its axis along \(Z\). In contrast, for \(Z^2/c^2-X^2/a^2-Y^2/b^2=1\), the positive \(Z^2\) term must be at least \(1\); hence \(|Z|\ge c\). This is the two-sheet form, with axis along \(Z\).
Worked Example: A One-Sheet Hyperboloid
Worked Example: A One-Sheet Hyperboloid
Classify \(4x^2+y^2-z^2=4\), identify its axis, and state a few points that can guide a corrected sketch.
Divide every term by \(4\):
There are two positive squared terms and one negative squared term. The negative term is the \(z\)-term, so this is a hyperboloid of one sheet with axis along \(z\). At \(z=0\), the equation reduces to \(x^2+y^2/4=1\); in particular, \((1,0,0)\) and \((0,2,0)\) are points on the surface. At \(x=y=0\), the equation gives \(-z^2/4=1\), which has no real solution.
Check the points in the original equation: for \((1,0,0)\), the left side is \(4\); for \((0,2,0)\), it is \(4\). Both equal the right side. The lack of points on the \(z\)-axis does not contradict the classification: the surface extends around that axis, rather than passing through it. A corrected sketch should show a connected surface oriented around the \(z\)-axis, with the ellipse at \(z=0\) as a useful reference.
Worked Example: A Sign Change Can Create or Remove Real Points
Worked Example: A Sign Change Can Create or Remove Real Points
Compare the two equations \(\frac{x^2}{4}+\frac{y^2}{9}+\frac{z^2}{16}=-1\) and \(\frac{x^2}{4}+\frac{y^2}{9}-\frac{z^2}{16}=-1\).
In the first equation, every term on the left is nonnegative, so the left side is at least \(0\). It cannot equal \(-1\). The equation has no real points. For example, even at the origin its left side is \(0\), not \(-1\); no choice of real \(x,y,z\) can make the sum negative.
In the second equation, rearrange by adding \(z^2/16\) to both sides and then multiplying the entire equation by \(-1\):
There is one positive squared term, the \(z\)-term, and two negative ones. This is a hyperboloid of two sheets with axis along \(z\). The points \((0,0,4)\) and \((0,0,-4)\) satisfy the normalized equation, as each gives \(16/16=1\). Substituting either point into the original equation gives \(-16/16=-1\), as required. In contrast, the origin gives \(0\) on the left of the original equation, so it is not on the surface. The corrected sketch has two separate parts, one on each side of the plane \(z=0\).
Worked Example: Completing Squares Without Losing a Sign
Worked Example: Completing Squares Without Losing a Sign
Classify and locate the surface \(\frac{x^2}{9}-\frac{4x}{9}+\frac{y^2}{4}+\frac{y}{2}-\frac{z^2}{16}+\frac{3z}{8}+\frac{19}{144}=1\).
Group terms by variable. For the \(x\)-terms, complete the square using \(x^2-4x=(x-2)^2-4\). For the \(y\)-terms, use \(y^2+2y=(y+1)^2-1\). For the \(z\)-terms, use \(z^2-6z=(z-3)^2-9\), remembering that the \(z\)-square has a negative coefficient:
The extra constants cancel because \(-4/9-1/4+9/16+19/144=(-64-36+81+19)/144=0\). Therefore the equation becomes
This is a hyperboloid of one sheet with axis parallel to the \(z\)-axis and center at \((2,-1,3)\). To check a point, take \((5,-1,3)\). In the completed-square form, the left side is \(9/9+0-0=1\). In the original equation, the left side is \(25/9-20/9+1/4-1/2-9/16+9/8+19/144\). With denominator \(144\), this is \((400-320+36-72-81+162+19)/144=144/144=1\). The point satisfies both forms.
A common error is to complete \(-z^2+6z\) as \(-(z-3)^2-9\). Expanding that expression gives \(-z^2+6z-18\), which is not \(-z^2+6z\). The correct identity is \(-z^2+6z=-(z-3)^2+9\). The sign on the added constant reverses because the whole square is being subtracted.
Worked Example: A Paraboloid Opens in the Direction Set by Its Sign
Worked Example: A Paraboloid Opens in the Direction Set by Its Sign
Determine the opening direction of \(\frac{x^2}{4}+\frac{y^2}{9}=-(z-1)\), and compare it with the equation obtained by changing the right side to \(z-1\).
Since both squared terms are nonnegative, solving for \(z\) gives
The vertex is \((0,0,1)\): substituting it into the original equation gives \(0=0\). At \((x,y)=(2,0)\), the equation gives \(z=1-1=0\); at \((x,y)=(0,3)\), it also gives \(z=1-1=0\). Checking either point in the original equation gives \(1=-(-1)\), so each lies on the surface. The surface opens in the negative \(z\)-direction from its vertex.
If the right side is changed to \(z-1\), then \(z=1+x^2/4+y^2/9\), so the surface opens in the positive \(z\)-direction instead. The vertex remains \((0,0,1)\), and the point \((2,0,2)\) satisfies the altered equation: its left side is \(1\), and its right side is \(2-1=1\). Changing this one sign reverses the opening direction.
A Reliable Sign-Checking Routine
Keep each squared variable with its own coefficient and linear term before completing squares.
If a squared term has a negative coefficient, factor that coefficient out first; it changes the sign of the constant adjustment.
Divide every term by the same nonzero constant. If multiplying the whole equation by \(-1\), change every sign, not just one.
Identify the relative signs, type, and axis or opening direction. Substitute a simple point into the original equation to check the algebra.
Common Mistake: Changing Only One Side
Suppose \(\frac{x^2}{4}+\frac{y^2}{9}-\frac{z^2}{16}=1\). Multiplying the entire equation by \(-1\) gives \(-\frac{x^2}{4}-\frac{y^2}{9}+\frac{z^2}{16}=-1\). These equations are equivalent: they have exactly the same solutions. If you multiply only the left side by \(-1\) and leave the right side unchanged, you get \(-\frac{x^2}{4}-\frac{y^2}{9}+\frac{z^2}{16}=1\), which is a different surface.
Likewise, moving a term across the equals sign changes its sign, but does not change the equation’s solution set. Changing a term’s sign without making the corresponding algebraic change is not a valid rearrangement. When a classification seems to contradict an intercept or a substituted point, return to the original equation and check each rearrangement line by line.
Why Correct Sketches Start with Correct Signs
The sign pattern is not a cosmetic detail on the page: it determines whether the equation describes a real surface, what type of quadric it describes, and which coordinate direction its axis or opening follows. Once the standard form is correct, its center or vertex and a few checked points help anchor the sketch. When those features conflict, suspect a sign error before drawing a new surface.
Check Your Understanding
Use the sign pattern first, then check any proposed classification by substitution or a brief algebraic argument.
- Classify \(\frac{x^2}{9}+\frac{y^2}{4}-\frac{z^2}{25}=1\) and identify its axis.
- Can \(\frac{x^2}{4}+\frac{y^2}{9}+\frac{z^2}{16}=-2\) have a real solution? Explain.
- What surface and axis are indicated by \(\frac{z^2}{9}-\frac{x^2}{4}-\frac{y^2}{16}=1\)?
- In \(-z^2+8z\), what is the correct completed-square form, including its constant adjustment?
- In which direction does \(\frac{x^2}{4}+\frac{y^2}{9}=-(z+2)\) open, and what is its vertex?