Question

Find the center, vertices, and foci of the ellipse with equation 2x2 + 8y2 = 16.

Answer 1

please help

Answer 2

\(\bf \cfrac{(x-h)^2}{a^2}+\cfrac{(y-k)^2}{b^2}=1\\
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2x^2 + 8y^2 = 16\\
\textit{divide both sides by 16}\\
\cfrac{2x^2}{16} + \cfrac{8y^2}{16} = \cfrac{16}{16} \implies \cfrac{x^2}{8} + \cfrac{y^2}{2} = 1 \implies \cfrac{(x-0)^2}{8} + \cfrac{(y-0)^2}{2} = 1\)

Answer 3

I don't have much time, but here's the idea. Your ellipse must be equal to 1. If its not equal to 1, then you must divide the whole equation by a number that will allow everything to be equal to 1. For you, you have 16 on the other side, meaning you must divide the whole equation by 16. Doing this gives us:
\[\frac{ x ^{2} }{ 8 }+\frac{ y ^{2} }{ 2 }=1\]So now we have the ellipse in the form of:
\[\frac{ x ^{2} }{ a ^{2} }+\frac{ y ^{2} }{ b ^{2} }\]The denominator which is the largest in an ellipse is always a^2. If a^2 is under the x, the ellipse stretches out along the x-axis. If the larger denominator is under y^2, then the ellipse stretches up the y-axis. So now for your verticies. From the center of your ellipse, the verticies are left and right a units and up and down b units. In order to get a and b, you take the square root of the two denominators. SO as of now, a^2 = 8, meaning a = 2sqrt(2). This tells us there are verticies 2sqrt(2) left and right of the center. b^2 = 2, meaning b equals sqrt(2). This tells us there are verticies up and down sqrt(2) units from the center. FInally, to find the foci, you must do this:
\[c=\sqrt{a ^{2}-b ^{2}} \]So when we solve that, we get that c is sqrt(6). Now the foci are stretched out in the same direction as the graph. So since this ellipse was stretched left and right along the x-axis, so were the verticies. The verticies are left and right sqrt(6) units. Sorry, dont really have time to wait for questions and fill in the gaps. Hope this helps some.

Answer 4

see the center? see the "a" and "b" components?

Answer 5

yes

Answer 6

so foci =(0,sqr6)???