# 8.2 The hyperbola  (Page 3/13)

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Given the equation of a hyperbola in standard form, locate its vertices and foci.

1. Determine whether the transverse axis lies on the x - or y -axis. Notice that $\text{\hspace{0.17em}}{a}^{2}\text{\hspace{0.17em}}$ is always under the variable with the positive coefficient. So, if you set the other variable equal to zero, you can easily find the intercepts. In the case where the hyperbola is centered at the origin, the intercepts coincide with the vertices.
1. If the equation has the form $\text{\hspace{0.17em}}\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}=1,$ then the transverse axis lies on the x -axis. The vertices are located at $\text{\hspace{0.17em}}\left(±a,0\right),$ and the foci are located at $\text{\hspace{0.17em}}\left(±c,0\right).$
2. If the equation has the form $\text{\hspace{0.17em}}\frac{{y}^{2}}{{a}^{2}}-\frac{{x}^{2}}{{b}^{2}}=1,$ then the transverse axis lies on the y -axis. The vertices are located at $\text{\hspace{0.17em}}\left(0,±a\right),$ and the foci are located at $\text{\hspace{0.17em}}\left(0,±c\right).$
2. Solve for $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ using the equation $\text{\hspace{0.17em}}a=\sqrt{{a}^{2}}.$
3. Solve for $\text{\hspace{0.17em}}c\text{\hspace{0.17em}}$ using the equation $\text{\hspace{0.17em}}c=\sqrt{{a}^{2}+{b}^{2}}.$

## Locating a hyperbola’s vertices and foci

Identify the vertices and foci of the hyperbola    with equation $\text{\hspace{0.17em}}\frac{{y}^{2}}{49}-\frac{{x}^{2}}{32}=1.$

The equation has the form $\text{\hspace{0.17em}}\frac{{y}^{2}}{{a}^{2}}-\frac{{x}^{2}}{{b}^{2}}=1,$ so the transverse axis lies on the y -axis. The hyperbola is centered at the origin, so the vertices serve as the y -intercepts of the graph. To find the vertices, set $\text{\hspace{0.17em}}x=0,$ and solve for $\text{\hspace{0.17em}}y.$

$\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}1=\frac{{y}^{2}}{49}-\frac{{x}^{2}}{32}\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}1=\frac{{y}^{2}}{49}-\frac{{0}^{2}}{32}\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}1=\frac{{y}^{2}}{49}\hfill \\ {y}^{2}=49\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}y=±\sqrt{49}=±7\hfill \end{array}$

The foci are located at $\text{\hspace{0.17em}}\left(0,±c\right).\text{\hspace{0.17em}}$ Solving for $\text{\hspace{0.17em}}c,$

$c=\sqrt{{a}^{2}+{b}^{2}}=\sqrt{49+32}=\sqrt{81}=9$

Therefore, the vertices are located at $\text{\hspace{0.17em}}\left(0,±7\right),$ and the foci are located at $\text{\hspace{0.17em}}\left(0,9\right).$

Identify the vertices and foci of the hyperbola with equation $\text{\hspace{0.17em}}\frac{{x}^{2}}{9}-\frac{{y}^{2}}{25}=1.$

Vertices: $\text{\hspace{0.17em}}\left(±3,0\right);\text{\hspace{0.17em}}$ Foci: $\text{\hspace{0.17em}}\left(±\sqrt{34},0\right)$

## Writing equations of hyperbolas in standard form

Just as with ellipses, writing the equation for a hyperbola in standard form allows us to calculate the key features: its center, vertices, co-vertices, foci, asymptotes, and the lengths and positions of the transverse and conjugate axes. Conversely, an equation for a hyperbola can be found given its key features. We begin by finding standard equations for hyperbolas centered at the origin. Then we will turn our attention to finding standard equations for hyperbolas centered at some point other than the origin.

## Hyperbolas centered at the origin

Reviewing the standard forms given for hyperbolas centered at $\text{\hspace{0.17em}}\left(0,0\right),$ we see that the vertices, co-vertices, and foci are related by the equation $\text{\hspace{0.17em}}{c}^{2}={a}^{2}+{b}^{2}.\text{\hspace{0.17em}}$ Note that this equation can also be rewritten as $\text{\hspace{0.17em}}{b}^{2}={c}^{2}-{a}^{2}.\text{\hspace{0.17em}}$ This relationship is used to write the equation for a hyperbola when given the coordinates of its foci and vertices.

Given the vertices and foci of a hyperbola centered at $\text{\hspace{0.17em}}\left(0,\text{\hspace{0.17em}}\text{0}\right),$ write its equation in standard form.

1. Determine whether the transverse axis lies on the x - or y -axis.
1. If the given coordinates of the vertices and foci have the form $\text{\hspace{0.17em}}\left(±a,0\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(±c,0\right),\text{\hspace{0.17em}}$ respectively, then the transverse axis is the x -axis. Use the standard form $\text{\hspace{0.17em}}\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}=1.$
2. If the given coordinates of the vertices and foci have the form $\text{\hspace{0.17em}}\left(0,±a\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(0,±c\right),\text{\hspace{0.17em}}$ respectively, then the transverse axis is the y -axis. Use the standard form $\text{\hspace{0.17em}}\frac{{y}^{2}}{{a}^{2}}-\frac{{x}^{2}}{{b}^{2}}=1.$
2. Find $\text{\hspace{0.17em}}{b}^{2}\text{\hspace{0.17em}}$ using the equation $\text{\hspace{0.17em}}{b}^{2}={c}^{2}-{a}^{2}.$
3. Substitute the values for $\text{\hspace{0.17em}}{a}^{2}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}{b}^{2}\text{\hspace{0.17em}}$ into the standard form of the equation determined in Step 1.

#### Questions & Answers

12, 17, 22.... 25th term
12, 17, 22.... 25th term
Akash
College algebra is really hard?
Absolutely, for me. My problems with math started in First grade...involving a nun Sister Anastasia, bad vision, talking & getting expelled from Catholic school. When it comes to math I just can't focus and all I can hear is our family silverware banging and clanging on the pink Formica table.
Carole
find the 15th term of the geometric sequince whose first is 18 and last term of 387
I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
hmm well what is the answer
Abhi
how do they get the third part x = (32)5/4
can someone help me with some logarithmic and exponential equations.
sure. what is your question?
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
hmm
Abhi
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Abhi
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Abhi
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salma
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salma
Commplementary angles
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Sherica
im all ears I need to learn
Sherica
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Tamia
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Uday
hi
salma
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
a perfect square v²+2v+_
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algebra 2 Inequalities:If equation 2 = 0 it is an open set?
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
rolling four fair dice and getting an even number an all four dice
Kristine 2*2*2=8