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

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0any ideas on the center of it?

xmistermayhem
 one year ago
Best ResponseYou've already chosen the best response.0I guess 0,0 ?? Not sure...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well, 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?

xmistermayhem
 one year ago
Best ResponseYou've already chosen the best response.03 to each foci, 1 is the minor axis?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeap.. so let's see now.. one sec

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is a vertical ellipse, meaing the major axis is over the yaxis, that is "a" is under the fraction with the "y" then so.... one sec

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so. the distance from the focus to the center, is "c" and \(\bf \cfrac{(x{\color{brown}{ h}})^2}{{\color{purple}{ b}}^2}+\cfrac{(y{\color{blue}{ k}})^2}{{\color{purple}{ a}}^2}=1 \qquad center\ ({\color{brown}{ h}},{\color{blue}{ k}})\qquad vertices\ ({\color{brown}{ h}}, {\color{blue}{ k}}\pm a) \\ \quad \\ \cfrac{(x{\color{brown}{ 0}})^2}{{\color{purple}{ 1}}^2}+\cfrac{(y{\color{blue}{ 0}})^2}{{\color{purple}{ a}}^2}=1 \qquad center\ ({\color{brown}{ 0}},{\color{blue}{ 0}}) \\ \quad \\ c=\sqrt{a^2b^2}\implies 3=\sqrt{a^21^2}\impliedby \textit{solve for "a"}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so anyhow \(\bf \cfrac{(x{\color{brown}{ h}})^2}{{\color{purple}{ b}}^2}+\cfrac{(y{\color{blue}{ k}})^2}{{\color{purple}{ a}}^2}=1 \qquad center\ ({\color{brown}{ h}},{\color{blue}{ k}})\qquad vertices\ ({\color{brown}{ h}}, {\color{blue}{ k}}\pm a) \\ \quad \\ \cfrac{(x{\color{brown}{ 0}})^2}{{\color{purple}{ 1}}^2}+\cfrac{(y{\color{blue}{ 0}})^2}{{\color{purple}{ a}}^2}=1 \qquad center\ ({\color{brown}{ 0}},{\color{blue}{ 0}}) \\ \quad \\ c=\sqrt{a^2b^2}\implies 3=\sqrt{a^21^2}\implies 3^2=a^21 \\ \quad \\ 9+1=a^2\implies \sqrt{10}=a\qquad thus \\ \quad \\ \cfrac{(x{\color{brown}{ 0}})^2}{{\color{purple}{ 1}}^2}+\cfrac{(y{\color{blue}{ 0}})^2}{{\color{purple}{ a}}^2}=1\implies \cfrac{x}{{\color{purple}{ 1}}^2}+\cfrac{y}{{\color{purple}{ (\sqrt{10})}}^2}=1 \\ \quad \\ \cfrac{x}{1}+\cfrac{y}{10}=1\implies x+\cfrac{y}{10}=1\)
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