Exercise 16 Page 825

We want to write an equation of a hyperbola with the given center, focus, and vertex. ccc Center & & Focus & & Vertex ( 6, 1) & & ( 6,6) & & ( 6,- 2) Since the vertex and the focus lie on the y-axis, we have a vertical hyperbola. Recall the main characteristics of this type of hyperbola.

We know that h = 6 and k = 1. From the given information, we can also deduce that a= - 3 and c= 5. To write the equation of the hyperbola, we need to find the value of b. To do so, we will substitute - 3 and 5 for a and c, respectively, into the equation c^2= a^2+ b^2.

c^2=a^2+b^2

SubstituteII

a= - 3, c= 5

5^2=( - 3)^2+b^2

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Solve for b

CalcPow

Calculate power

25=9+b^2

SubEqn

LHS-9=RHS-9

16=b^2

SqrtEqn

sqrt(LHS)=sqrt(RHS)

4=b

RearrangeEqn

Rearrange equation

b=4

Note that when solving for b, we were able to only consider the principal root because the value of b is squared in the equation of a hyperbola. Therefore, its sign is not relevant. Now that we know that b= 4, we can write the equation of the hyperbola. (y- 1)^2/( - 3)^2 - (x- 6)^2/4^2 = 1 ⇕ (y-1)^2/9-(x-6)^2/16=1 Next, let's graph the hyperbola. To do so, let's first isolate y in our equation.

(y-1)^2/9-(x-6)^2/16=1

AddEqn

LHS+(x-6)^2/16=RHS+(x-6)^2/16

(y-1)^2/9=1+(x-6)^2/16

SqrtEqn

sqrt(LHS)=sqrt(RHS)

y-1/3= ± sqrt(1+(x-6)^2/16)

MultEqn

LHS * 3=RHS* 3

y-1 = ± 3sqrt(1+(x-6)^2/16)

AddEqn

LHS+1=RHS+1

y = ± 3sqrt(1+(x-6)^2/16) + 1

Next, let's make a table of values.

x	± 3sqrt(1+(x-6)^2/16) + 1	y = 3sqrt(1+(x-6)^2/16) + 1	y = -3sqrt(1+(x-6)^2/16) + 1
- 1	± 3sqrt(1+( - 1-6)^2/16) + 1	≈ 7	≈ - 5
3	± 3sqrt(1+( 3-6)^2/16) + 1	4.75	- 2.75
6	± 3sqrt(1+( 6-6)^2/16) + 1	4	- 2
9	± 3sqrt(1+( 9-6)^2/16) + 1	4.75	- 2.75
13	± 3sqrt(1+( 13-6)^2/16) + 1	≈ 7	≈ - 5

Now let's plot the points and connect each set of points with a smooth curve.