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The standard form of the equation of an ellipse with center $\text{\hspace{0.17em}}\left(h,\text{}k\right)\text{\hspace{0.17em}}$ and major axis parallel to the x -axis is
where
The standard form of the equation of an ellipse with center $\text{\hspace{0.17em}}\left(h,k\right)\text{\hspace{0.17em}}$ and major axis parallel to the y -axis is
where
Just as with ellipses centered at the origin, ellipses that are centered at a point $\text{\hspace{0.17em}}\left(h,k\right)\text{\hspace{0.17em}}$ have vertices, co-vertices, and foci that are related by the equation $\text{\hspace{0.17em}}{c}^{2}={a}^{2}-{b}^{2}.\text{\hspace{0.17em}}$ 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.
Given the vertices and foci of an ellipse not centered at the origin, write its equation in standard form.
What is the standard form equation of the ellipse that has vertices $\text{\hspace{0.17em}}\left(\mathrm{-2},\mathrm{-8}\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(\mathrm{-2},\text{2}\right)$
and foci $\text{\hspace{0.17em}}\left(\mathrm{-2},\mathrm{-7}\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(\mathrm{-2},\text{1}\right)?$
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
First, we identify the center, $\text{\hspace{0.17em}}\left(h,k\right).\text{\hspace{0.17em}}$ The center is halfway between the vertices, $\text{\hspace{0.17em}}\left(-\mathrm{2,}-8\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(-2,\text{2}\right).\text{\hspace{0.17em}}$ Applying the midpoint formula, we have:
Next, we find $\text{\hspace{0.17em}}{a}^{2}.\text{\hspace{0.17em}}$ The length of the major axis, $\text{\hspace{0.17em}}2a,$ is bounded by the vertices. We solve for $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ by finding the distance between the y -coordinates of the vertices.
So $\text{\hspace{0.17em}}{a}^{2}=25.$
Now we find $\text{\hspace{0.17em}}{c}^{2}.\text{\hspace{0.17em}}$ The foci are given by $\text{\hspace{0.17em}}\left(h,k\pm c\right).\text{\hspace{0.17em}}$ So, $\text{\hspace{0.17em}}\left(h,k-c\right)=\left(\mathrm{-2},\mathrm{-7}\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(h,k+c\right)=\left(\mathrm{-2},\text{1}\right).\text{\hspace{0.17em}}$ We substitute $\text{\hspace{0.17em}}k=\mathrm{-3}\text{\hspace{0.17em}}$ using either of these points to solve for $\text{\hspace{0.17em}}c.$
So $\text{\hspace{0.17em}}{c}^{2}=16.$
Next, we solve for $\text{\hspace{0.17em}}{b}^{2}\text{\hspace{0.17em}}$ using the equation $\text{\hspace{0.17em}}{c}^{2}={a}^{2}-{b}^{2}.$
Finally, we substitute the values found for $\text{\hspace{0.17em}}h,k,{a}^{2},$ and $\text{\hspace{0.17em}}{b}^{2}\text{\hspace{0.17em}}$ into the standard form equation for an ellipse:
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