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anonymous
 one year ago
Write the equation of a hyperbola with a center at (5, 3), vertices at (5, 5) and (5, 1) and covertices at (11, 3) and (1, 3)
anonymous
 one year ago
Write the equation of a hyperbola with a center at (5, 3), vertices at (5, 5) and (5, 1) and covertices at (11, 3) and (1, 3)

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well, since the ycoordinate is what changes between the two vertices, this will be a hyperbola opening up and down, which means the fraction that has the yvariable will be the positive of the two fractions in the equation. So we know we have a hyperbola of this form \[\frac{ (yk)^{2} }{ a^{2} }  \frac{ (xh)^{2} }{ b^{2} } = 1\] The a^2 is always with the positive fraction and the b^2 is always with the negative fraction when dealing with a hyperbola. Now we are given the center (h,k) = (5,3). So we can plug those values in and have this now: \[\frac{ (y+3)^{2} }{ a^{2} }  \frac{ (x+5)^{2} }{ b^{2} } = 1\] Now, the denominator of the fraction with the variable x represents a distance left and right from the center while the denominator of the fraction with y in it reprsents a distance up and down from the center. Since each vertex is up 2 or down 2 from the center, this means a = 2 and thus a^2 = 4. Now I havent heard the term covertices used before, but I assume it just means the values that would be on the imaginary box that is normally used to sketch the asymptotes. Either way, this distance is left 6 and right 6 from the center, therefore we have b = 6 which means b^2 = 36. Thus our full equation is \[\frac{ (y+3)^{2} }{ 4 }  \frac{ (x+5)^{2} }{ 36 } = 1\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Concentrationalizing

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well, its center is (3,2) based on the equation, so the ellipse will be symmetric about its center. So yes

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Concentrationalizing thanks how about this one

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well, you need to know which denominator would give you vertices and which one would give you covertices. So first off, which equations are ellipses and which ones are hyperbolas?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well, I explain a lot of it to ya, but there are things you will eventually have to pick up from studying and such. An ellipse has an equation of this form: \[\frac{ (xh)^{2} }{ a^{2} } + \frac{ (yk)^{2} }{ b^{2} } = 1\] The order of a^2 and b^2 may be flipflopped depending on the equation A hyperbola has an equation of this form: \[\frac{ (xh)^{2} }{ a^{2} }  \frac{ (yk)^{2} }{ b^{2} }=1\] Which fraction contains x and which one contains y may be flipflopped in a hyperbola So can you see which ones would be ellipses and which would be hyperbolas?
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