Question

What is the equation of the ellipse with foci (0, 3), (0, -3) and co-vertices (1, 0), (-1, 0)?

Answer 1

any ideas on the center of it?

Answer 2

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

Answer 3

well, let's see


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?

Answer 4

3 to each foci, 1 is the minor axis?

Answer 5

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

Answer 6

Oki doki

Answer 7

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

Answer 8

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"}\)

Answer 9

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\)