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

• This Question is Open
1. anonymous

@ganeshie8

2. anonymous

@TheSmartOne

3. anonymous

Well, since the y-coordinate is what changes between the two vertices, this will be a hyperbola opening up and down, which means the fraction that has the y-variable will be the positive of the two fractions in the equation. So we know we have a hyperbola of this form $\frac{ (y-k)^{2} }{ a^{2} } - \frac{ (x-h)^{2} }{ b^{2} } = 1$ 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 co-vertices 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 $\frac{ (y+3)^{2} }{ 4 } - \frac{ (x+5)^{2} }{ 36 } = 1$

4. anonymous

@Concentrationalizing

5. anonymous

Well, its center is (3,2) based on the equation, so the ellipse will be symmetric about its center. So yes

6. anonymous

7. anonymous

Well, you need to know which denominator would give you vertices and which one would give you co-vertices. So first off, which equations are ellipses and which ones are hyperbolas?

8. anonymous

i have no clue

9. anonymous

Well, 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: $\frac{ (x-h)^{2} }{ a^{2} } + \frac{ (y-k)^{2} }{ b^{2} } = 1$ The order of a^2 and b^2 may be flip-flopped depending on the equation A hyperbola has an equation of this form: $\frac{ (x-h)^{2} }{ a^{2} } - \frac{ (y-k)^{2} }{ b^{2} }=1$ Which fraction contains x and which one contains y may be flip-flopped in a hyperbola So can you see which ones would be ellipses and which would be hyperbolas?