# Foci: (0,0), (0,8), major axis of length 16

Question
Solid Geometry
Foci: (0,0), (0,8), major axis of length 16

2020-10-29
The foci are in the vertical direction so the ellipse is vertical. Its equation must then be of the form $$\displaystyle{\left(\frac{{\left({x}−{h}\right)}^{{2}}}{{b}^{{2}}}\right)}+{\left(\frac{{\left({y}−{k}\right)}^{{2}}}{{a}^{{2}}}\right)}={1}.$$
The center of the ellipse is the midpoint between the foci. Since the foci are (0,0) and (0,8), the center is (h,k)=((0+0)/2,(0+8)/2)=(0/2,8/2)=(0,4). The distance from (0,4) to (0,0) is 4 so c=4
The length of the major axis of an ellipse is 2a so the length of the major axis is 16, then 2a=16. Dividing both sides by 2 then gives a=8.
The values of aa, bb, and cc are related by the equation c2=a2−b2. Substitute the values of a=8 and c=4 into this equation to find b2:
$$\displaystyle{c}^{{2}}={a}^{{2}}-{b}^{{2}}$$
$$\displaystyle{4}^{{2}}={8}^{{2}}-{b}^{{2}}$$
$$\displaystyle{16}={64}-{b}^{{2}}$$
$$\displaystyle-{48}=-{b}^{{2}}$$
$$\displaystyle{48}={b}^{{2}}$$
Substituting in h=0, k=4, a=8 and b2=48 then gives:
$$\displaystyle{\left(\frac{{\left({x}-{h}\right)}^{{2}}}{{b}^{{2}}}\right)}+\frac{{{\left({y}-{k}\right)}^{{2}}}}{{a}^{{2}}}={1}$$
$$\displaystyle{\left(\frac{{\left({x}-{0}\right)}^{{2}}}{{48}}\right)}+\frac{{{\left({y}-{4}\right)}^{{2}}}}{{8}^{{2}}}={1}$$
$$\displaystyle{\left({x}^{{2}}+{48}\right)}+\frac{{{\left({y}-{4}\right)}^{{2}}}}{{64}}={1}$$

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