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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.
Let's start by recalling the equation of hyperbolas centered at the origin.
ccc
Horizontal & & Vertical
Hyperbola & & Hyperbola
x^2/a^2-y^2/b^2=1 & & y^2/a^2-x^2/b^2=1
Consider the given equation.
| 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 |
Using this information we can identify that the vertices are (0,± sqrt(18)) = (0,± 3sqrt(2)). Let's substitute a=3sqrt(2) and b=sqrt(20)=2sqrt(5) into the formula for the asymptotes and obtain their equations.
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
The asymptotes are y=± 3sqrt(10)10x. Now let's calculate c, the absolute value of the nonzero coordinate of the foci. To do so we will substitute a=3sqrt(2) and b=2sqrt(5) into c^2=a^2+b^2.
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)
Note that when solving the above equation we only needed to consider the principal root because c is a positive number. The foci of the hyperbola are (0,± sqrt(38)).
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!