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The equation of a horizontal hyperbola is x^2a^2- y^2b^2=1. The vertices are (± a,0). How can you find the foci and the asymptotes?
Vertices: (± 6, 0)
Foci: (± sqrt(37),0)
Asymptotes: y=± 1/6x
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.
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Horizontal & & Vertical
Hyperbola & & Hyperbola
x^2/a^2-y^2/b^2=1 & & y^2/a^2-x^2/b^2=1
Now we will rewrite the given equation to match one of these formats.
.LHS /36.=.RHS /36.
Write as a difference of fractions
a/b=.a /36./.b /36.
Write as a power
From the above formula we can see that the equation represents a horizontal hyperbola. Next, let's review the main characteristics of this type of hyperbola.
| Horizontal Hyperbola with Center (0,0) | |
|---|---|
| Equation | x^2/a^2-y^2/b^2=1 |
| Transverse axis | Horizontal |
| Vertices | (± a,0) |
| Foci | (± c,0), where c^2= a^2+ b^2 |
| Asymptotes | y=± b/ax |
Using this information, we can identify that the vertices are ( ± 6,0 ). Let's substitute a=6 and b=1 into the formula for the asymptotes and obtain their equations.
The asymptotes are y=± 16x. Now let's calculate c, the absolute value of the nonzero coordinate of the foci. To do so we will substitute a=6 and b=1 into c^2=a^2+b^2.
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 (± sqrt(37),0).
To graph the function let's summarize all of the information that we have found.
| Equation | x^2/6^2-y^2/1^2=1 |
| Transverse axis | Horizontal |
| Vertices | (± 6,0) |
| Foci | (± sqrt(37),0) |
| Asymptotes | y=± 1/6x |
Finally, we can graph our hyperbola!