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The equation of a vertical hyperbola is y^2a^2- x^2b^2=1. The vertices are (0,± a). How can you find the foci and the asymptotes?
Vertices: (0,± 3sqrt(2))
Foci: (0,± sqrt(38))
Asymptotes: y=± 3sqrt(10)/10x
Graph:
We will find the desired information and use it to draw the graph of the hyperbola.
y^2/18-x^2/20=1 ⇔ y^2/( sqrt(18))^2-x^2/( sqrt(20))^2=1 From the above formula we can see that the equation represents a vertical hyperbola. Next, let's review the main characteristics of this type of hyperbola.
| Vertical Hyperbola with Center (0,0) | |
|---|---|
| Equation | y^2/a^2-x^2/b^2=1 |
| Transverse axis | Vertical |
| Vertices | (0,± a) |
| Foci | (0,± c), where c^2= a^2+ b^2 |
| Asymptotes | y=± a/bx |
a= 3sqrt(2), b= 2sqrt(5)
a/b=a * sqrt(5)/b * sqrt(5)
sqrt(a)* sqrt(a)= a
sqrt(a)*sqrt(b)=sqrt(a* b)
Multiply
a= 3sqrt(2), b= 2sqrt(5)
(a * b)^m=a^m* b^m
( sqrt(a) )^2 = a
Calculate power and product
Add terms
sqrt(LHS)=sqrt(RHS)
To graph the function let's summarize all of the information that we have found.
| Equation | y^2/(3sqrt(2))^2-x^2/(2sqrt(5))^2=1 |
| Transverse axis | Vertical |
| Vertices | (0,± 3sqrt(2)) |
| Foci | (0, ± sqrt(38)) |
| Asymptotes | y=± 3sqrt(10)/10x |
Finally, we can graph our hyperbola!