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Start by identifying whether the hyperbola is vertical or horizontal.
x^2/36-y^2/28=1
We are given the graph of a hyperbola and we want to write its equation.
This is a vertical hyperbola. Let's recall the general equation for this type of conic section and some of its characteristics.
| Equation | (y- k)^2/a^2-(x- h)^2/b^2=1 |
|---|---|
| Center | ( h, k) |
| Foci | ( h, k± c) |
| Vertices | ( h, k± a) |
We see in the diagram that the vertices are (0, 6) and (0, - 6). With these values we can find the length of the transverse axis, 2 a.
To find the center we will use the fact that it is the midpoint between the vertices. ( 0+0/2, 6+(-6)/2) ⇔ ( 0, 0) Therefore, we have that h= 0 and k= 0. To find the value of b we will consider the foci of the hyperbola. We see in the diagram that the distance from the center to the foci is 8. Therefore, we have that c = 8. With this information we can set up and solve an equation.
Substitute values
Calculate power
LHS-36=RHS-36
sqrt(LHS)=sqrt(RHS)
Rearrange equation
Note that since b>0 we took the principal root in the above calculation. We can finally write the equation of the hyperbola. (x- 0)^2/6^2-(y- 0)^2/(sqrt(28))^2=1 ⇕ x^2/36-y^2/28=1