## xmistermayhem one year ago What is the equation of the ellipse with foci (0, 3), (0, -3) and co-vertices (1, 0), (-1, 0)?

1. anonymous

any ideas on the center of it?

2. xmistermayhem

I guess 0,0 ?? Not sure...

3. anonymous

well, let's see |dw:1441153187872:dw| so... yes, is at the origin, or 0,0 well. what's the distance from the center to either focus? what about the minor axis, what's the length of the minor axis?

4. xmistermayhem

3 to each foci, 1 is the minor axis?

5. anonymous

yeap.. so let's see now.. one sec

6. xmistermayhem

Oki doki

7. anonymous

is a vertical ellipse, meaing the major axis is over the y-axis, that is "a" is under the fraction with the "y" then so.... one sec

8. anonymous

so. the distance from the focus to the center, is "c" and $$\bf \cfrac{(x-{\color{brown}{ h}})^2}{{\color{purple}{ b}}^2}+\cfrac{(y-{\color{blue}{ k}})^2}{{\color{purple}{ a}}^2}=1 \qquad center\ ({\color{brown}{ h}},{\color{blue}{ k}})\qquad vertices\ ({\color{brown}{ h}}, {\color{blue}{ k}}\pm a) \\ \quad \\ \cfrac{(x-{\color{brown}{ 0}})^2}{{\color{purple}{ 1}}^2}+\cfrac{(y-{\color{blue}{ 0}})^2}{{\color{purple}{ a}}^2}=1 \qquad center\ ({\color{brown}{ 0}},{\color{blue}{ 0}}) \\ \quad \\ c=\sqrt{a^2-b^2}\implies 3=\sqrt{a^2-1^2}\impliedby \textit{solve for "a"}$$

9. anonymous

so anyhow $$\bf \cfrac{(x-{\color{brown}{ h}})^2}{{\color{purple}{ b}}^2}+\cfrac{(y-{\color{blue}{ k}})^2}{{\color{purple}{ a}}^2}=1 \qquad center\ ({\color{brown}{ h}},{\color{blue}{ k}})\qquad vertices\ ({\color{brown}{ h}}, {\color{blue}{ k}}\pm a) \\ \quad \\ \cfrac{(x-{\color{brown}{ 0}})^2}{{\color{purple}{ 1}}^2}+\cfrac{(y-{\color{blue}{ 0}})^2}{{\color{purple}{ a}}^2}=1 \qquad center\ ({\color{brown}{ 0}},{\color{blue}{ 0}}) \\ \quad \\ c=\sqrt{a^2-b^2}\implies 3=\sqrt{a^2-1^2}\implies 3^2=a^2-1 \\ \quad \\ 9+1=a^2\implies \sqrt{10}=a\qquad thus \\ \quad \\ \cfrac{(x-{\color{brown}{ 0}})^2}{{\color{purple}{ 1}}^2}+\cfrac{(y-{\color{blue}{ 0}})^2}{{\color{purple}{ a}}^2}=1\implies \cfrac{x}{{\color{purple}{ 1}}^2}+\cfrac{y}{{\color{purple}{ (\sqrt{10})}}^2}=1 \\ \quad \\ \cfrac{x}{1}+\cfrac{y}{10}=1\implies x+\cfrac{y}{10}=1$$

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