McGraw Hill Glencoe Algebra 2, 2012
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McGraw Hill Glencoe Algebra 2, 2012 View details
2. Verifying Trigonometric Identities
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Exercise 70 Page 885

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:

Practice makes perfect

We will find the desired information and use it to draw the graph of the hyperbola.

Vertices, Foci, and Asymptotes

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.

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
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.
y=± a/bx
y=± 3sqrt(2)/2sqrt(5)x
Simplify
y=± 3sqrt(2)* sqrt(5)/2sqrt(5)* sqrt(5)x
y=± 3sqrt(2)* sqrt(5)/2* 5x
y=± 3sqrt(10)/2* 5x
y=± 3sqrt(10)/10x
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.
c^2=a^2+b^2
c^2=( 3sqrt(2))^2+( 2sqrt(5))^2
Solve for c
c^2=3^2 (sqrt(2))^2+2^2(sqrt(5))^2
c^2=3^2(2)+2^2(5)
c^2=18+20
c^2=38
c= sqrt(38)
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)).

Graph

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!