Exercise 1 Page 668

We want to find the equation that represents the graph of a hyperbola with foci at

(5, 0)

and

(- 5, 0) .

Because the foci are in the

(\pm c, 0)

form, this gives us that that the equation is the first type of standard form for hyperbolas, where the transverse axis is horizontal with vertices

(\pm a, 0) .

\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1

With this knowledge we can eliminate B and D as the answer, because they are not in the same standard form. Remember the equation that relates the general hyperbola and its foci.

c^{2} = a^{2} + b^{2}

Now, let's use the

a

and

b

value in the equation of choice A and solve for

c .

This way we can compare it the given

c

values in our problem.

\frac{x ^{2}}{2 5} - \frac{y ^{2}}{4} = 1

We can substitute in our values,

a^{2} = 25

and

b^{2} = 4,

to find

c .

c^{2} = a^{2} + b^{2}

Substitute

$a^{2} = 25$ and $b^{2} = 4$

c^{2} = 25 + 4

AddTerms

Add terms

c^{2} = 29

SqrtEqn

$LHS = RHS$

c = \pm 29

This is not equal to our foci at

(5, 0)

and

(- 5, 0),

which means our answer is C. We can follow the same steps using those

a

and

b

values to double check.

\frac{x ^{2}}{2 1} - \frac{y ^{2}}{4} = 1