Writing Equations of Ellipses Not Centered at the Origin
Like the graphs of other equations, the graph of an ellipse can be translated. If an ellipse is translated [latex]h[/latex] units horizontally and [latex]k[/latex] units vertically, the center of the ellipse will be [latex]\left(h,k\right)[/latex]. This translation results in the standard form of the equation we saw previously, with [latex]x[/latex] replaced by [latex]\left(x-h\right)[/latex] and y replaced by [latex]\left(y-k\right)[/latex].
A General Note: Standard Forms of the Equation of an Ellipse with Center (h, k)
The standard form of the equation of an ellipse with center [latex]\left(h,\text{ }k\right)[/latex] and major axis parallel to the x-axis is[latex]\frac{{\left(x-h\right)}^{2}}{{a}^{2}}+\frac{{\left(y-k\right)}^{2}}{{b}^{2}}=1[/latex]
where
- [latex]a>b[/latex]
- the length of the major axis is [latex]2a[/latex]
- the coordinates of the vertices are [latex]\left(h\pm a,k\right)[/latex]
- the length of the minor axis is [latex]2b[/latex]
- the coordinates of the co-vertices are [latex]\left(h,k\pm b\right)[/latex]
- the coordinates of the foci are [latex]\left(h\pm c,k\right)[/latex], where [latex]{c}^{2}={a}^{2}-{b}^{2}[/latex].
[latex]\frac{{\left(x-h\right)}^{2}}{{b}^{2}}+\frac{{\left(y-k\right)}^{2}}{{a}^{2}}=1[/latex]
where
- [latex]a>b[/latex]
- the length of the major axis is [latex]2a[/latex]
- the coordinates of the vertices are [latex]\left(h,k\pm a\right)[/latex]
- the length of the minor axis is [latex]2b[/latex]
- the coordinates of the co-vertices are [latex]\left(h\pm b,k\right)[/latex]
- the coordinates of the foci are [latex]\left(h,k\pm c\right)[/latex], where [latex]{c}^{2}={a}^{2}-{b}^{2}[/latex].
Figure 7. (a) Horizontal ellipse with center [latex]\left(h,k\right)[/latex] (b) Vertical ellipse with center [latex]\left(h,k\right)[/latex]How To: Given the vertices and foci of an ellipse not centered at the origin, write its equation in standard form.
- Determine whether the major axis is parallel to the x- or y-axis.
- If the y-coordinates of the given vertices and foci are the same, then the major axis is parallel to the x-axis. Use the standard form [latex]\frac{{\left(x-h\right)}^{2}}{{a}^{2}}+\frac{{\left(y-k\right)}^{2}}{{b}^{2}}=1[/latex].
- If the x-coordinates of the given vertices and foci are the same, then the major axis is parallel to the y-axis. Use the standard form [latex]\frac{{\left(x-h\right)}^{2}}{{b}^{2}}+\frac{{\left(y-k\right)}^{2}}{{a}^{2}}=1[/latex].
- Identify the center of the ellipse [latex]\left(h,k\right)[/latex] using the midpoint formula and the given coordinates for the vertices.
- Find [latex]{a}^{2}[/latex] by solving for the length of the major axis, [latex]2a[/latex], which is the distance between the given vertices.
- Find [latex]{c}^{2}[/latex] using [latex]h[/latex] and [latex]k[/latex], found in Step 2, along with the given coordinates for the foci.
- Solve for [latex]{b}^{2}[/latex] using the equation [latex]{c}^{2}={a}^{2}-{b}^{2}[/latex].
- Substitute the values for [latex]h,k,{a}^{2}[/latex], and [latex]{b}^{2}[/latex] into the standard form of the equation determined in Step 1.
Example 2: Writing the Equation of an Ellipse Centered at a Point Other Than the Origin
What is the standard form equation of the ellipse that has vertices [latex]\left(-2,-8\right)[/latex] and [latex]\left(-2,\text{2}\right)[/latex] and foci [latex]\left(-2,-7\right)[/latex] and [latex]\left(-2,\text{1}\right)?[/latex]Solution
The x-coordinates of the vertices and foci are the same, so the major axis is parallel to the y-axis. Thus, the equation of the ellipse will have the form[latex]\frac{{\left(x-h\right)}^{2}}{{b}^{2}}+\frac{{\left(y-k\right)}^{2}}{{a}^{2}}=1[/latex]
First, we identify the center, [latex]\left(h,k\right)[/latex]. The center is halfway between the vertices, [latex]\left(-2,-8\right)[/latex] and [latex]\left(-2,\text{2}\right)[/latex]. Applying the midpoint formula, we have:
[latex]\begin{array}{l}\left(h,k\right)=\left(\frac{-2+\left(-2\right)}{2},\frac{-8+2}{2}\right)\hfill \\ \text{ }=\left(-2,-3\right)\hfill \end{array}[/latex]
Next, we find [latex]{a}^{2}[/latex]. The length of the major axis, [latex]2a[/latex], is bounded by the vertices. We solve for [latex]a[/latex] by finding the distance between the y-coordinates of the vertices.
[latex]\begin{array}{c}2a=2-\left(-8\right)\\ 2a=10\\ a=5\end{array}[/latex]
So [latex]{a}^{2}=25[/latex].
Now we find [latex]{c}^{2}[/latex]. The foci are given by [latex]\left(h,k\pm c\right)[/latex]. So, [latex]\left(h,k-c\right)=\left(-2,-7\right)[/latex] and [latex]\left(h,k+c\right)=\left(-2,\text{1}\right)[/latex]. We substitute [latex]k=-3[/latex] using either of these points to solve for [latex]c[/latex].
[latex]\begin{array}{c}k+c=1\\ -3+c=1\\ c=4\end{array}[/latex]
So [latex]{c}^{2}=16[/latex].
Next, we solve for [latex]{b}^{2}[/latex] using the equation [latex]{c}^{2}={a}^{2}-{b}^{2}[/latex].
[latex]\begin{array}{c}{c}^{2}={a}^{2}-{b}^{2}\\ 16=25-{b}^{2}\\ {b}^{2}=9\end{array}[/latex]
Finally, we substitute the values found for [latex]h,k,{a}^{2}[/latex], and [latex]{b}^{2}[/latex] into the standard form equation for an ellipse:
[latex]\frac{{\left(x+2\right)}^{2}}{9}+\frac{{\left(y+3\right)}^{2}}{25}=1[/latex]
Try It 2
What is the standard form equation of the ellipse that has vertices [latex]\left(-3,3\right)[/latex] and [latex]\left(5,3\right)[/latex] and foci [latex]\left(1 - 2\sqrt{3},3\right)[/latex] and [latex]\left(1+2\sqrt{3},3\right)?[/latex] SolutionLicenses & Attributions
CC licensed content, Specific attribution
- Precalculus. Provided by: OpenStax Authored by: OpenStax College. Located at: https://cnx.org/contents/[email protected]:1/Preface. License: CC BY: Attribution.
