xmistermayhem
  • xmistermayhem
What is the equation of the ellipse with foci (0, 3), (0, -3) and co-vertices (1, 0), (-1, 0)?
Mathematics
schrodinger
  • schrodinger
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

jdoe0001
  • jdoe0001
any ideas on the center of it?
xmistermayhem
  • xmistermayhem
I guess 0,0 ?? Not sure...
jdoe0001
  • jdoe0001
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?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

xmistermayhem
  • xmistermayhem
3 to each foci, 1 is the minor axis?
jdoe0001
  • jdoe0001
yeap.. so let's see now.. one sec
xmistermayhem
  • xmistermayhem
Oki doki
jdoe0001
  • jdoe0001
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
jdoe0001
  • jdoe0001
so. the distance from the focus to the center, is "c" and \(\bf \cfrac{(x-{\color{brown}{ h}})^2}{{\color{purple}{ b}}^2}+\cfrac{(y-{\color{blue}{ k}})^2}{{\color{purple}{ a}}^2}=1 \qquad center\ ({\color{brown}{ h}},{\color{blue}{ k}})\qquad vertices\ ({\color{brown}{ h}}, {\color{blue}{ k}}\pm a) \\ \quad \\ \cfrac{(x-{\color{brown}{ 0}})^2}{{\color{purple}{ 1}}^2}+\cfrac{(y-{\color{blue}{ 0}})^2}{{\color{purple}{ a}}^2}=1 \qquad center\ ({\color{brown}{ 0}},{\color{blue}{ 0}}) \\ \quad \\ c=\sqrt{a^2-b^2}\implies 3=\sqrt{a^2-1^2}\impliedby \textit{solve for "a"}\)
jdoe0001
  • jdoe0001
so anyhow \(\bf \cfrac{(x-{\color{brown}{ h}})^2}{{\color{purple}{ b}}^2}+\cfrac{(y-{\color{blue}{ k}})^2}{{\color{purple}{ a}}^2}=1 \qquad center\ ({\color{brown}{ h}},{\color{blue}{ k}})\qquad vertices\ ({\color{brown}{ h}}, {\color{blue}{ k}}\pm a) \\ \quad \\ \cfrac{(x-{\color{brown}{ 0}})^2}{{\color{purple}{ 1}}^2}+\cfrac{(y-{\color{blue}{ 0}})^2}{{\color{purple}{ a}}^2}=1 \qquad center\ ({\color{brown}{ 0}},{\color{blue}{ 0}}) \\ \quad \\ c=\sqrt{a^2-b^2}\implies 3=\sqrt{a^2-1^2}\implies 3^2=a^2-1 \\ \quad \\ 9+1=a^2\implies \sqrt{10}=a\qquad thus \\ \quad \\ \cfrac{(x-{\color{brown}{ 0}})^2}{{\color{purple}{ 1}}^2}+\cfrac{(y-{\color{blue}{ 0}})^2}{{\color{purple}{ a}}^2}=1\implies \cfrac{x}{{\color{purple}{ 1}}^2}+\cfrac{y}{{\color{purple}{ (\sqrt{10})}}^2}=1 \\ \quad \\ \cfrac{x}{1}+\cfrac{y}{10}=1\implies x+\cfrac{y}{10}=1\)

Looking for something else?

Not the answer you are looking for? Search for more explanations.