We've updated our
Privacy Policy effective December 15. Please read our updated Privacy Policy and tap

Study Guides > College Algebra

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 hh units horizontally and kk units vertically, the center of the ellipse will be (h,k)\left(h,k\right). This translation results in the standard form of the equation we saw previously, with xx replaced by (xh)\left(x-h\right) and y replaced by (yk)\left(y-k\right).

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 (h, k)\left(h,\text{ }k\right) and major axis parallel to the x-axis is
(xh)2a2+(yk)2b2=1\frac{{\left(x-h\right)}^{2}}{{a}^{2}}+\frac{{\left(y-k\right)}^{2}}{{b}^{2}}=1
where
  • a>ba>b
  • the length of the major axis is 2a2a
  • the coordinates of the vertices are (h±a,k)\left(h\pm a,k\right)
  • the length of the minor axis is 2b2b
  • the coordinates of the co-vertices are (h,k±b)\left(h,k\pm b\right)
  • the coordinates of the foci are (h±c,k)\left(h\pm c,k\right), where c2=a2b2{c}^{2}={a}^{2}-{b}^{2}.
The standard form of the equation of an ellipse with center (h,k)\left(h,k\right) and major axis parallel to the y-axis is
(xh)2b2+(yk)2a2=1\frac{{\left(x-h\right)}^{2}}{{b}^{2}}+\frac{{\left(y-k\right)}^{2}}{{a}^{2}}=1
where
  • a>ba>b
  • the length of the major axis is 2a2a
  • the coordinates of the vertices are (h,k±a)\left(h,k\pm a\right)
  • the length of the minor axis is 2b2b
  • the coordinates of the co-vertices are (h±b,k)\left(h\pm b,k\right)
  • the coordinates of the foci are (h,k±c)\left(h,k\pm c\right), where c2=a2b2{c}^{2}={a}^{2}-{b}^{2}.
Just as with ellipses centered at the origin, ellipses that are centered at a point (h,k)\left(h,k\right) have vertices, co-vertices, and foci that are related by the equation c2=a2b2{c}^{2}={a}^{2}-{b}^{2}. We can use this relationship along with the midpoint and distance formulas to find the equation of the ellipse in standard form when the vertices and foci are given.
(a) Horizontal ellipse with center (h,k) (b) Vertical ellipse with center (h,k) Figure 7. (a) Horizontal ellipse with center (h,k)\left(h,k\right) (b) Vertical ellipse with center (h,k)\left(h,k\right)

How To: Given the vertices and foci of an ellipse not centered at the origin, write its equation in standard form.

  1. Determine whether the major axis is parallel to the x- or y-axis.
    1. 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 (xh)2a2+(yk)2b2=1\frac{{\left(x-h\right)}^{2}}{{a}^{2}}+\frac{{\left(y-k\right)}^{2}}{{b}^{2}}=1.
    2. 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 (xh)2b2+(yk)2a2=1\frac{{\left(x-h\right)}^{2}}{{b}^{2}}+\frac{{\left(y-k\right)}^{2}}{{a}^{2}}=1.
  2. Identify the center of the ellipse (h,k)\left(h,k\right) using the midpoint formula and the given coordinates for the vertices.
  3. Find a2{a}^{2} by solving for the length of the major axis, 2a2a, which is the distance between the given vertices.
  4. Find c2{c}^{2} using hh and kk, found in Step 2, along with the given coordinates for the foci.
  5. Solve for b2{b}^{2} using the equation c2=a2b2{c}^{2}={a}^{2}-{b}^{2}.
  6. Substitute the values for h,k,a2h,k,{a}^{2}, and b2{b}^{2} 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 (2,8)\left(-2,-8\right) and (2,2)\left(-2,\text{2}\right) and foci (2,7)\left(-2,-7\right) and (2,1)?\left(-2,\text{1}\right)?

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
(xh)2b2+(yk)2a2=1\frac{{\left(x-h\right)}^{2}}{{b}^{2}}+\frac{{\left(y-k\right)}^{2}}{{a}^{2}}=1
First, we identify the center, (h,k)\left(h,k\right). The center is halfway between the vertices, (2,8)\left(-2,-8\right) and (2,2)\left(-2,\text{2}\right). Applying the midpoint formula, we have:
(h,k)=(2+(2)2,8+22) =(2,3)\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}
Next, we find a2{a}^{2}. The length of the major axis, 2a2a, is bounded by the vertices. We solve for aa by finding the distance between the y-coordinates of the vertices.
2a=2(8)2a=10a=5\begin{array}{c}2a=2-\left(-8\right)\\ 2a=10\\ a=5\end{array}
So a2=25{a}^{2}=25. Now we find c2{c}^{2}. The foci are given by (h,k±c)\left(h,k\pm c\right). So, (h,kc)=(2,7)\left(h,k-c\right)=\left(-2,-7\right) and (h,k+c)=(2,1)\left(h,k+c\right)=\left(-2,\text{1}\right). We substitute k=3k=-3 using either of these points to solve for cc.
k+c=13+c=1c=4\begin{array}{c}k+c=1\\ -3+c=1\\ c=4\end{array}
So c2=16{c}^{2}=16. Next, we solve for b2{b}^{2} using the equation c2=a2b2{c}^{2}={a}^{2}-{b}^{2}.
c2=a2b216=25b2b2=9\begin{array}{c}{c}^{2}={a}^{2}-{b}^{2}\\ 16=25-{b}^{2}\\ {b}^{2}=9\end{array}
Finally, we substitute the values found for h,k,a2h,k,{a}^{2}, and b2{b}^{2} into the standard form equation for an ellipse:
(x+2)29+(y+3)225=1\frac{{\left(x+2\right)}^{2}}{9}+\frac{{\left(y+3\right)}^{2}}{25}=1

Try It 2

What is the standard form equation of the ellipse that has vertices (3,3)\left(-3,3\right) and (5,3)\left(5,3\right) and foci (123,3)\left(1 - 2\sqrt{3},3\right) and (1+23,3)?\left(1+2\sqrt{3},3\right)? Solution

Licenses & Attributions