Question

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)

Answer 1

@ganeshie8

Answer 2

@TheSmartOne

Answer 3

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
\[\frac{ (y+3)^{2} }{ 4 } - \frac{ (x+5)^{2} }{ 36 } = 1\]

Answer 4

@Concentrationalizing

Answer 5

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

Answer 6

@Concentrationalizing thanks how about this one

Answer 7

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?

Answer 8

i have no clue

Answer 9

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?