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Coordinates and Quadric Surfaces · Tutorial 18 of 999

How Translations and Scaling Change a Quadric

See how translations and scale changes alter the position and dimensions of standard spheres and quadric surfaces.

Beginner 9 min read

What You'll Learn

  • Read the center or vertex directly from translated coordinate expressions.
  • Interpret squared denominators as squared length scales in standard forms.
  • Distinguish changing a surface’s size from multiplying its entire equation by a constant.
  • Predict how translations and scale changes affect a quadric’s sketch.
  • Check a proposed point by substituting it into the equation.

Position and Size Are Separate Features

A standard quadric equation carries information about both where a surface sits and how large it is. Expressions such as \(x-h\) shift the surface’s position; squared denominators set length scales; and constants determine the level set being described. Reading these features separately makes it easier to compare equations and anticipate the effect on a sketch.

For a sphere, the pattern is familiar: changing the center moves the sphere, while changing the radius changes its size. The same distinction applies to ellipsoids, hyperboloids, cones, and paraboloids, although a paraboloid has a vertex rather than a center. In each case, a translated coordinate expression identifies a reference location, and the remaining terms determine how the surface extends away from it.

Key idea: A translation changes a quadric’s location without changing its shape or size. A change to its squared length parameters changes its scale. A constant on the right-hand side must be interpreted in the context of the whole equation.

Translations: Moving the Reference Point

In three dimensions, translating a point by \((h,k,\ell)\) means adding those coordinate changes to its original coordinates. In a quadric equation, the corresponding translated coordinate expressions are \(x-h\), \(y-k\), and \(z-\ell\). They measure displacement from the reference point \((h,k,\ell)\).

For example, the sphere equation

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

describes a sphere centered at the origin with radius \(3\). Replacing the coordinate expressions by \(x+2\), \(y-1\), and \(z-3\) gives

$$ (x+2)^2+(y-1)^2+(z-3)^2=9. $$

This is still a sphere of radius \(3\), but its center is \((-2,1,3)\). The sign inside each parentheses is opposite to the corresponding center coordinate: \(x+2=x-(-2)\).

For a paraboloid, the same idea locates the vertex. For instance, \((x-1)^2+(y+2)^2=z-3\) has its vertex at \((1,-2,3)\). It does not have a center; the point identifies where the surface begins and from which its cross-sections extend.

Squared Denominators: Setting Length Scales

Consider the ellipsoid

$$ \frac{x^2}{4}+\frac{y^2}{9}+\frac{z^2}{16}=1. $$

The denominators are the squares of the three semi-axis lengths. The lengths are therefore \(2\), \(3\), and \(4\), not \(4\), \(9\), and \(16\). The ellipsoid extends twice as far from its center in the \(x\)-direction as in the \(y\)-direction, and four times as far in the \(z\)-direction as in the \(x\)-direction.

For a hyperboloid or cone, denominators also set scale parameters, but the signs determine which standard form is present. Changing a denominator does not change a plus sign into a minus sign. For a paraboloid, the coefficients and denominators determine how rapidly the surface widens as it moves away from its vertex.

Interpretation: A denominator \(a^2\) corresponds to a length scale \(a\). Multiplying that denominator by \(4\) doubles the corresponding length scale, because \(\sqrt{4a^2}=2a\).

Why the Translation and Scale Rules Work

Both rules follow by expressing a point on the surface relative to a reference point. Let \((u,v,w)\) be coordinates measured from the origin of a standard-form equation, and let the reference location in the sketch be \((h,k,\ell)\). A translation places the point with relative coordinates \((u,v,w)\) at \((h+u,k+v,\ell+w)\). Consequently, the relative coordinates of a point \((x,y,z)\) are \(u=x-h\), \(v=y-k\), and \(w=z-\ell\).

Now suppose each relative coordinate is also scaled by a positive factor: \(s_x\) in the \(x\)-direction, \(s_y\) in the \(y\)-direction, and \(s_z\) in the \(z\)-direction. Substituting \(u=(x-h)/s_x\), \(v=(y-k)/s_y\), and \(w=(z-\ell)/s_z\) into a standard ellipsoid gives the translated and scaled equation below.

$$ \frac{u^2}{a^2}+\frac{v^2}{b^2}+\frac{w^2}{c^2}=1 \quad\Longrightarrow\quad \frac{(x-h)^2}{(s_xa)^2} +\frac{(y-k)^2}{(s_yb)^2} +\frac{(z-\ell)^2}{(s_zc)^2}=1. $$

The resulting semi-axis lengths are \(s_xa\), \(s_yb\), and \(s_zc\). The reference point changes through \((h,k,\ell)\); each length scale changes through its own factor. This substitution argument also explains why the same location-and-scale reading applies to standard forms with different sign patterns: translation changes the coordinate expressions, while scaling changes the squared length parameters.

1
Identify the reference point.
Read \(h\), \(k\), and \(\ell\) from expressions of the form \(x-h\), \(y-k\), and \(z-\ell\).
2
Read the scale parameters.
Take the square roots of the squared denominators to obtain the associated length scales.
3
Keep the sign pattern in view.
Translations and positive changes to denominators do not change the plus and minus signs that distinguish standard forms.

Worked Examples: Reading Paired Equations

Worked Example: Translating and Enlarging a Sphere

Compare the origin-centered sphere \(x^2+y^2+z^2=9\) with the sphere

$$ (x+2)^2+(y-1)^2+(z-3)^2=36. $$

The first sphere has center \((0,0,0)\) and radius \(\sqrt{9}=3\). In the second equation, \(x+2=x-(-2)\), \(y-1=y-1\), and \(z-3=z-3\). Its center is therefore \((-2,1,3)\). Its radius is \(\sqrt{36}=6\). Relative to the first sphere, it has moved to a new center and doubled in radius.

Sketch cue: Place the second center at \((-2,1,3)\), then imagine a sphere reaching twice as far in every direction from that point as the first sphere does from the origin.

Check: The point \((-2,1,9)\) is six units above the center. Substitution gives

$$ (-2+2)^2+(1-1)^2+(9-3)^2=0+0+36=36. $$

It lies on the sphere, as expected for a point six units from its center.

Worked Example: Translating and Rescaling an Ellipsoid

Compare the ellipsoid

$$ \frac{x^2}{4}+\frac{y^2}{9}+\frac{z^2}{16}=1 $$

with

$$ \frac{(x-1)^2}{25}+\frac{(y+2)^2}{9}+\frac{(z-3)^2}{4}=1. $$

The first ellipsoid has center \((0,0,0)\) and semi-axis lengths \(\sqrt{4}=2\), \(\sqrt{9}=3\), and \(\sqrt{16}=4\). The second has center \((1,-2,3)\), since \(y+2=y-(-2)\). Its semi-axis lengths are \(\sqrt{25}=5\), \(\sqrt{9}=3\), and \(\sqrt{4}=2\). Thus its \(x\)-scale has increased, its \(y\)-scale is unchanged, and its \(z\)-scale has decreased. Its principal directions remain parallel to the coordinate axes.

Sketch cue: Move the ellipsoid’s center from the origin to \((1,-2,3)\). Draw it longer in the \(x\)-direction than the original, equally extended in the \(y\)-direction, and shorter in the \(z\)-direction.

Check: The point \((6,-2,3)\) is five units from the center in the \(x\)-direction, so it should lie on the surface. Substitution confirms this:

$$ \frac{(6-1)^2}{25}+\frac{(-2+2)^2}{9}+\frac{(3-3)^2}{4} =\frac{25}{25}+0+0=1. $$

Worked Example: Scaling a Hyperboloid of One Sheet

Consider the pair

$$ \frac{x^2}{4}+\frac{y^2}{1}-\frac{z^2}{9}=1 \qquad\text{and}\qquad \frac{(x-2)^2}{9}+\frac{(y+1)^2}{4}-\frac{(z-3)^2}{16}=1. $$

In the first equation, the reference point is the origin, and the scale parameters are \(2\), \(1\), and \(3\). In the second, the center is \((2,-1,3)\), and the scale parameters are \(3\), \(2\), and \(4\). The plus and minus signs are unchanged, so the second equation retains the same one-sheet type and orientation: the positive terms are in the \(x\)- and \(y\)-directions, and the negative term is in the \(z\)-direction. The changes move its center and alter its dimensional scales; they do not interchange its sign pattern.

Sketch cue: Use the same general one-sheet form as for the first equation, but place its center at \((2,-1,3)\) and use the new scales along the coordinate directions.

Check: Set \(y=-1\), \(z=3\), and \(x=5\). These coordinates are three units along the \(x\)-direction from the center. Substitution into the second equation gives

$$ \frac{(5-2)^2}{9}+\frac{(-1+1)^2}{4}-\frac{(3-3)^2}{16} =\frac{9}{9}+0-0=1. $$

The point satisfies the equation.

Worked Example: Moving and Scaling a Paraboloid

Compare \(x^2+y^2=z\) with

$$ \frac{(x-1)^2}{4}+\frac{(y+2)^2}{9}=\frac{z-3}{6}. $$

The original paraboloid has vertex \((0,0,0)\). The second has vertex \((1,-2,3)\). At a vertical displacement \(z-3=6\), the right-hand side is \(1\), so the cross-section satisfies \((x-1)^2/4+(y+2)^2/9=1\). Its semi-axis lengths are \(2\) and \(3\). For comparison, the original equation at \(z=1\) gives \(x^2+y^2=1\), a circular cross-section of radius \(1\). The second surface has been translated, and its widening has different horizontal scales.

Sketch cue: Put the vertex at \((1,-2,3)\); the surface opens in the positive \(z\)-direction, with wider cross-sections in the \(y\)-direction than in the \(x\)-direction.

Check: The point \((3,-2,9)\) is two units from the vertex in \(x\) and six units above it. The left and right sides are

$$ \frac{(3-1)^2}{4}+\frac{(-2+2)^2}{9} =\frac{4}{4}+0=1, \qquad \frac{9-3}{6}=\frac{6}{6}=1. $$

Both sides agree, so the point lies on the surface.

A Common Mistake: Confusing a Denominator with a Length

A frequent error is to read the denominator \(25\) in \((x-1)^2/25\) as a length of \(25\). The denominator is a squared length parameter, so the corresponding length is \(\sqrt{25}=5\). The same mistake can make a surface appear far larger or smaller than its equation describes.

A separate issue arises when an entire equation is multiplied by a nonzero constant. For example, \(x^2+y^2+z^2=9\) and \(2x^2+2y^2+2z^2=18\) describe the same sphere: dividing the second equation by \(2\) recovers the first. Multiplying every term on both sides by the same nonzero number does not change the solution set. By contrast, changing \(x^2+y^2+z^2=9\) to \(x^2+y^2+z^2=18\) changes the sphere’s radius from \(3\) to \(\sqrt{18}=3\sqrt{2}\).

Keep the two operations distinct: Multiplying an entire equation by a nonzero constant preserves its surface. Changing only a level constant or a squared denominator can change the surface’s size.

Using the Rules in a Sketch

When comparing two standard-form equations, first locate their centers or vertices, then compare the squared denominators or other scale parameters. The position and scale changes can be read independently, while the sign pattern remains visible in the terms. Paired equations are particularly useful: they show whether a surface has merely moved, changed size, or done both.

These readings are predictions about the sketch, not substitutes for checking the equation. When a point is proposed as lying on a surface, substitution verifies that claim directly. For a broader sketch, keep each equation’s standard form intact and use its reference location and scales to place and size the surface consistently.

Check Your Understanding

Use the equation features and scale interpretations developed above to answer these questions.

  1. What are the center and radius of \((x-3)^2+(y+1)^2+(z-4)^2=49\)?
  2. What semi-axis lengths correspond to denominators \(36\), \(4\), and \(25\) in an ellipsoid’s standard equation?
  3. How does \(x^2/4+y^2/9-z^2/16=1\) differ in position and scale from \((x-1)^2/4+(y+2)^2/9-(z-3)^2/16=1\)?
  4. Why do \(x^2+y^2+z^2=9\) and \(5x^2+5y^2+5z^2=45\) describe the same surface?
  5. For \((x-2)^2/9+(y+1)^2/4=(z-5)/3\), identify the vertex and the horizontal semi-axis lengths at \(z=8\).