Eccentricity is a characteristic that determines the shape of the conic sections. The eccentricity of the hyperbolas depends on the length of the transverse axis and the length of the conjugate axis. On the other hand, the length of these axes can be found directly in the equation of a hyperbola or using the coordinates of the vertices and the foci.
Here, we will learn about the equation that we can use to find the value of the eccentricity of hyperbolas. Then, we will look at some practice problems.
PRECALCULUS
Relevant for…
Learning about the eccentricity of the hyperbola with examples.
PRECALCULUS
Relevant for…
Learning about the eccentricity of the hyperbola with examples.
How to calculate the eccentricity of hyperbolas?
The eccentricity of a hyperbola is calculated using the length of the transverse axis and the length of the conjugate axis. Recall that hyperbolas are defined as the set of points in a plane, which have a difference of the distances from the two foci that is constant.
This results in the eccentricity of the hyperbolas always being greater than 1, that is, we have $latex e>1$.
The general equation of a hyperbola is:
$latex \frac{{{x}^2}}{{{a}^2}}-\frac{{{y}^2}}{{{b}^2}}=1$
The eccentricity formula is:
$latex e=\sqrt{1+\frac{{{b}^2}}{{{a}^2}}}$
where a is the length of the conjugate axis and b is the length of the transverse axis. The conjugate axis is the segment that connects the vertices and the transverse axis is the segment that connects the covertices.
Additionally, we can also calculate the eccentricity of a hyperbola, using the following formula:
$latex e=\frac{c}{a}$
where c is the length of the segment connecting the foci and can be calculated using the equation $latex {{c}^2}={{a}^2}+{{b}^2}$.
Eccentricity of a hyperbola – Examples with answers
The following examples are solved using the formula for the eccentricity of hyperbolas. Each example has its respective answer, but it is recommended that you try to solve them yourself before looking at the solution.
EXAMPLE 1
If a hyperbola has the equation $latex \frac{{{x}^2}}{25}-\frac{{{y}^2}}{16}=1$, what is its eccentricity?
Solution
We see that the equation has the form $latex \frac{{{x}^2}}{{{a}^2}}-\frac{{{y}^2}}{{{b}^2}}=1$, then, we can obtain the following values:
$latex {{a}^2}=25$
$latex {{b}^2}=16$
Now, we use these values in the eccentricity formula:
$latex e=\sqrt{1+\frac{{{b}^2}}{{{a}^2}}}$
$latex e=\sqrt{1+\frac{16}{25}}$
$latex e=1.28$
EXAMPLE 2
We have the hyperbola $latex \frac{{{x}^2}}{64}-\frac{{{y}^2}}{25}=1$. What is its eccentricity?
Solution
Again, we have an equation with the same form as the previous one, so we can recognize the following values:
$latex {{a}^2}=64$
$latex {{b}^2}=25$
Using these values in the eccentricity formula, we have:
$latex e=\sqrt{1+\frac{{{b}^2}}{{{a}^2}}}$
$latex e=\sqrt{1+\frac{25}{64}}$
$latex e=1.18$
EXAMPLE 3
What is the eccentricity of the hyperbola $latex \frac{{{x}^2}}{121}-\frac{{{y}^2}}{81}=1$?
Solution
We have to determine the values of $latex {{a}^2}$ and $latex {{b}^2}$. Therefore, we have:
$latex {{a}^2}=121$
$latex {{b}^2}=81$
We use the eccentricity formula with these values:
$latex e=\sqrt{1+\frac{{{b}^2}}{{{a}^2}}}$
$latex e=\sqrt{1+\frac{81}{121}}$
$latex e=1.2$
EXAMPLE 4
If a hyperbola has the equation $latex \frac{{{x}^2}}{120}-\frac{{{y}^2}}{40}=1$, what is its eccentricity?
Solution
We recognize the following values by looking at the equation of the hyperbola:
$latex {{a}^2}=120$
$latex {{b}^2}=40$
Now, we replace these values in the eccentricity formula and solve:
$latex e=\sqrt{1+\frac{{{b}^2}}{{{a}^2}}}$
$latex e=\sqrt{1+\frac{40}{120}}$
$latex e=1.15$
Eccentricity of a hyperbola – Practice problems
Practice using the formula for the eccentricity of a hyperbola by solving the following problems. Select an answer and check it to see if you got the correct answer.
See also
Interested in learning more about hyperbolas? Take a look at these pages:
Jefferson Huera Guzman
Jefferson is the lead author and administrator of Neurochispas.com. The interactive Mathematics and Physics content that I have created has helped many students.
