Shifted Conics
A messy equation like x squared plus y squared minus 6x plus 4y minus 12 equals 0 hides a perfectly ordinary circle. Completing the square in x and in y rewrites that general second-degree equation in a standard form whose center, vertices, and shape can be read off directly.
By the end of this page you can:
- Recognize that a parabola, ellipse, or hyperbola keeps its shape when its center or vertex moves to a point (h, k).
- Complete the square in x and in y to convert a general second-degree equation into a standard shifted form.
- Read the center, vertices, and orientation of the shifted conic from that form and graph it.
The floor. A shifted conic is a standard conic translated to a new center (h, k). Group the x terms, group the y terms, complete the square on each, and the standard form appears.
Quick Reference
| Field | Detail |
|---|---|
| Textbook | Stewart, Redlin, Watson 7e |
| Chapter | 11 |
| Section | 11.4 Shifted Conics |
| Open alternate | OpenStax Precalculus 2e, 12.4 |
| Difficulty | core |
| Time | 30 minutes |
Before You Start
The Idea
Replacing x with x minus h and y with y minus k in any standard conic equation slides the whole curve so its center (or vertex) sits at the point (h, k) instead of the origin. The shape, size, and orientation stay the same; only the location changes.
A general second-degree equation with no xy term can be sorted into x terms and y terms, and each group can be turned into a perfect square. The result is a shifted standard form. Which conic appears depends on the squared terms: matching positive coefficients give a circle or ellipse, opposite signs give a hyperbola, and a single squared variable gives a parabola.
Key Formulas
A conic centered or with vertex at (h, k) is obtained by the translation x maps to x minus h and y maps to y minus k. The shifted standard forms are:
\[ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \quad \text{(ellipse, center } (h,k)) \]
\[ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \quad \text{(hyperbola, center } (h,k)) \]
\[ (x-h)^2 = 4p\,(y-k) \quad \text{(parabola, vertex } (h,k)) \]
Source: Stewart 7e, Section 11.4 Shifted Conics.
The conversion tool is completing the square:
\[ x^2 + bx = \left(x + \tfrac{b}{2}\right)^2 - \left(\tfrac{b}{2}\right)^2 \]
Source: Stewart 7e, Section 11.4 Shifted Conics.
Worked Example
Identify and graph the conic given by
\[ x^2 + y^2 - 6x + 4y - 12 = 0. \]
Group the x terms and the y terms, and move the constant to the right side:
\[ (x^2 - 6x) + (y^2 + 4y) = 12. \]
Complete the square on each group. For x, half of negative 6 is negative 3, and its square is 9. For y, half of 4 is 2, and its square is 4. Add both to each side:
\[ (x^2 - 6x + 9) + (y^2 + 4y + 4) = 12 + 9 + 4. \]
Show the answer
The squares fold up and the right side adds to 25:
\[ (x-3)^2 + (y+2)^2 = 25. \]
The matching coefficients of one each identify a circle. Its center is (3, negative 2) and its radius is the square root of 25, which is 5. To graph it, plot the center at (3, negative 2) and draw the circle through the points 5 units left, right, up, and down from the center.
Connections
This idea builds directly on the standard parabola, the standard ellipse, and the standard hyperbola, since a shifted conic is one of those curves moved off the origin. It also rests on completing the square, the algebra step that does the conversion. Reading the center and orientation of a shifted conic prepares you for analyzing general second-degree equations and, in later courses, for rotating axes to handle conics whose axes are tilted.
Resources
- Stewart, Redlin, Watson, Precalculus 7e, Section 11.4 (primary).
- OpenStax Precalculus 2e, Section 12.4 (alternate).
Last updated: 2026-06-24