Question

The foci of an ellipse are located at (0,6) and (0,-6). The sum of the distances from any point on the ellipse to the foci is 20. Determine the equation of the ellipse. A) x2 36 + y2 64 = 1 B) x2 100 + y2 64 = 1 C) x2 64 + y2 100 = 1 D) x2 36 + y2 100 = 1

Answer 1

check the picture below.

we know the distances from a point on the ellipse to the foci is 20, namely as you see in the picture, if we use a point and run a line to one of the foci, then run another line from the same point to the other foci, add those two distances and we get 20.

in the picture, notice, we picked a point, then move it up some, and move up some more, the sum of those two legs/sides is always 20, so if we keep on moving that dot all the way till it's lined up collinearly with the foci, that distance is 20 still.

now, the foci are 6 units from the center, as you see there, so from focus to focus is 6+6 or 12 units, so from the upper focus to the upper vertex is 8 units, because 20 - 12 = 8.

all that means that, from the center at the origin, a = 6+8 = 14, as you see in the picture, and of course c = 6.

`\bf \textit{ellipse, vertical major axis} \\\\ \cfrac{(x- h)^2}{ b^2}+\cfrac{(y- k)^2}{ a^2}=1 \qquad \begin{cases} center\ ( h, k)\\ vertices\ ( h, k\pm a)\\ c=\textit{distance from}\\ \qquad \textit{center to foci}\\ \qquad √( a ^2- b ^2) \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}`

`\bf \begin{cases} h=0\\ k=0\\ a=14\\ c=6 \end{cases}\implies \cfrac{(x-0)^2}{b^2}+\cfrac{(y-0)^2}{14^2}=1 \\\\[-0.35em] ~\dotfill\\\\ c=√(a^2-b^2)\implies 6=√(14^2-b^2)\implies 6^2=14^2-b^2 \\\\\\ b^2=14^2-6^2\implies b^2=160\implies b=√(160) \\\\[-0.35em] ~\dotfill\\\\ \cfrac{(x-0)^2}{(√(160))^2}+\cfrac{(y-0)^2}{14^2}=1\implies \cfrac{x^2}{160}+\cfrac{y^2}{196}=1`