Question

Ellipse question please help

Answer 1

Check the picture below, so the ellipse looks more or less like so.

now, the center is clearly at the origin, and since the foci are one above and the other below, we can pretty much see is a vertical ellipse, the major axis is vertical.

we know it has an eccentricity of 1/2, BUT eccentricity "c" distance over "a" distance, but we already know from the distances of the foci that c = 3, how can it be 1/2? well, we can simply provide an equivalent fraction with "2" :)

`\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)\\ eccentricity\quad e=\cfrac{c}{a} \end{cases} \\\\[-0.35em] ~\dotfill`

`\begin{cases} h=0\\ k=0\\ c=3\\ e=(1)/(2)\to (3)/(6) \end{cases}\hspace{5em}e=\cfrac{\stackrel{c}{3}}{\underset{a}{6}}\impliedby \textit{this means a = 6} \\\\\\ c=√(a^2-b^2)\implies 3=√(6^2-b^2)\implies 9=36-b^2 \\\\\\ 9+b^2=36\implies b^2=27 \\\\[-0.35em] ~\dotfill\\\\ \cfrac{(x- 0)^2}{ 27}+\cfrac{(y- 0)^2}{ 6^2}=1\implies {\Large \begin{array}{llll} \cfrac{x^2}{27}+\cfrac{y^2}{36}=1 \end{array}}`