Exercise 49 Page 649

The equation of a horizontal hyperbola is x^2a^2- y^2b^2=1. The vertices are (0,± a). How can you find the foci and the asymptotes?

Vertices: (- 3,0) and (3,0)
Foci: (sqrt(13),0 ) and (- sqrt(13),0 )
Asymptotes: y=± 2/3x
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 Now we will rewrite the given equation to match one of these formats.

4x^2-9y^2=36

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Simplify left-hand side

DivEqn

.LHS /36.=.RHS /36.

x^2/9-y^2/4=1

WritePow

Write as a power

x^2/3^2-y^2/2^2=1

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 (± 3,0). Let's substitute a=3, and b=2 into the formula for the asymptotes and obtain their equations.

y=± b/ax

SubstituteII

a= 3, b= 2

y=± 2/3x

The asymptotes are y=± 23x. Now, let's calculate c. To do so, we will substitute a=3 and b=2 into c^2=a^2+b^2.

c^2=a^2+b^2

SubstituteII

a= 3, b= 2

c^2=3^2+2^2

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Solve for c

CalcPow