IXL - Find the eccentricity of a hyperbola (Precalculus practice)

The graph shows a hyperbola. What is its eccentricity?

Write your answer in simplified, rationalized form.

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question

The graph shows a hyperbola. What is its eccentricity?

Write your answer in simplified, rationalized form.

key idea

The equation of a hyperbola that has a vertical orientation in standard form is:

(

–

)

–

(

–

)

The center is

(

)

the vertices are

(

)

and the co-vertices are

(

)

Note that the vertices are on the hyperbola, but the co-vertices are not.

The transverse axis is the line segment with endpoints at the vertices and has length

The conjugate axis is the line segment with endpoints at the co-vertices and has length

(

)

(

–

)

(

)

(

–

)

(

)

The eccentricity is

where

is the distance from the center to a focus satisfying

The eccentricity of a hyperbola is always greater than 1.

solution

The eccentricity cannot be found by simply looking at a graph. So, first find the center. Then, find

and

Finally, calculate the length

and find

Find the center.

The asymptotes intersect at the center of the hyperbola.

The center is

(

)

Find

and

The hyperbola opens up and down, so it has a vertical orientation. This means the transverse axis is vertical and the conjugate axis is horizontal.

The length of the transverse axis is the distance between the two vertices, which is 12. Since

is half the length of the transverse axis,

The length of the conjugate axis is the distance between the two co-vertices, which is 6. Since

is half the length of the conjugate axis,

Calculate

then take the square root.

Since

is a distance, it is positive. So,

Find

The eccentricity is:

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