Question

an ellipse has a center of (3,1), a focus of (8,1) and a directrix of x = 36.8. what is the equation of the ellipse?

Answer 1

hmmm was going to do a quick mockup, but it might be too wide, so let's nevermind that piece.

so let's see, the center of this ellipse is at (3,1), and a focus point is at (8,1), notice, the y-coordinate is the same, meaning the distance between those two points is "c", which is 5, so c = 5.

We also know that the directrix is x = 36.8, well, that's a vertical line, if the line is vertical, that means the ellipse is running perpendicular to that, namely it has a major axis over the x-axis, so is horizontal.

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

`\bf \stackrel{\textit{its directrix}}{x=h\pm\cfrac{a^2}{c}}\implies \stackrel{\textit{using one directrix}}{36.8=h+\cfrac{a^2}{c}}\qquad \begin{cases} h=3\\ k=1\\ c=5 \end{cases}\implies 36.8=3+\cfrac{a^2}{5} \\\\\\ 33.8=\cfrac{a^2}{5}\implies \boxed{169=a^2}\implies 13=a \\\\[-0.35em] ~\dotfill`

`\bf \stackrel{\textit{recall that }}{c=√(a^2-b^2)}\implies 5=√(169-b^2)\implies 5^2=169-b^2\implies 5^2+b^2=169 \\\\\\ b^2=169-5^2\implies \boxed{b^2=144}\implies b=12 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \cfrac{(x-3)^2}{169}+\cfrac{(y-1)^2}{144}=1~\hfill`

Answer 2

a on edgn btw

Explanation: