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On this page, we discuss hyperbolas in rectangular coordinates. For a discussion of hyperbolas in polar form, see this separate page.

Hyperbola

The hyperbola is the most complicated of the three and, consequently, the most interesting. Figure 1 contains information about the hyperbola.
The vertices (\(\pm a\) on this plot) and the foci (F1 and F2 on this plot) lie on the transverse axis with the center at \((h,k)\). The standard equations are

\(\displaystyle{ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 }\)

horizontal transverse axis

\(\displaystyle{ \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 }\)

vertical transverse axis

As we did with the ellipse, we define an intermediate value \(c^2=a^2+b^2\) which will help us locate the foci.
For a hyperbola, we need to know the equations of the lines in blue on the plot. These are asymptotes.
For a horizontal transverse axis, the asymptotes are \(\displaystyle{ y=k \pm \frac{b}{a}(x-h) }\).
For a vertical transverse axis, the asymptotes are \(\displaystyle{ y=k \pm \frac{a}{b}(x-h) }\).
Similar to the ellipse, we define the eccentricity as \(e=c/a\). The results are summarized next.

\(\displaystyle{ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 }\)

horizontal transverse axis

center

\((h,k)\)

vertices

\((h \pm a, k)\)

foci

\( (h \pm c,k) \)

asymptotes

\(\displaystyle{ y=k \pm \frac{b}{a}(x-h) }\)

\(c^2=a^2+b^2\)

eccentricity

\( e=c/a \)

\(\displaystyle{ \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 }\)

vertical transverse axis

center

\((h,k)\)

vertices

\((h, k \pm a)\)

foci

\( (h,k \pm c) \)

asymptotes

\(\displaystyle{ y=k \pm \frac{a}{b}(x-h) }\)

\(c^2=a^2+b^2\)

eccentricity

\( e=c/a \)

Notes

1. Notice that the value of c is different here than for an ellipse.
2. Since \(c > a\), \(e > 1\).

The hyperbola is quite a complicated graph with lots of features. Figure 2 gives you an idea of some of the other features that are involved with hyperbolas. Here is a video that goes into more detail.

MIP4U - Conic Sections: The Hyperbola part 1 of 2 [4min-12secs]

video by MIP4U

Okay, time for some practice problems on hyperbolas. Although it may seem like we have given you a lot of information on this page, we have just skimmed the surface in the discussion of these three figures, especially the hyperbola. There are a lot more interesting features that we hope you get to explore in your class.

Practice

Sketch the graph of the hyperbola \(\displaystyle{ (y+4)^2 - \frac{x^2}{25} = 1 }\).

Problem Statement

Sketch the graph of the hyperbola \(\displaystyle{ (y+4)^2 - \frac{x^2}{25} = 1 }\).

Solution

1592 solution video

video by PatrickJMT

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Graph \(\displaystyle{ \frac{x^2}{49} - \frac{y^2}{25} = 1 }\).

Problem Statement

Graph \(\displaystyle{ \frac{x^2}{49} - \frac{y^2}{25} = 1 }\).

Solution

1604 solution video

video by PatrickJMT

close solution

Write the equation of the hyperbola that has vertices \((-2,-5), (4,-5)\) and foci \((-4,-5), (6,-5)\).

Problem Statement

Write the equation of the hyperbola that has vertices \((-2,-5), (4,-5)\) and foci \((-4,-5), (6,-5)\).

Solution

1605 solution video

video by PatrickJMT

close solution

Graph \(\displaystyle{ \frac{(x-2)^2}{4} - \frac{(y+3)^2}{9} = 1 }\).

Problem Statement

Graph \(\displaystyle{ \frac{(x-2)^2}{4} - \frac{(y+3)^2}{9} = 1 }\).

Solution

1606 solution video

video by MIP4U

close solution

Graph \(\displaystyle{ \frac{(y+4)^2}{4} - \frac{(x-2)^2}{16} = 1 }\).

Problem Statement

Graph \(\displaystyle{ \frac{(y+4)^2}{4} - \frac{(x-2)^2}{16} = 1 }\).

Solution

1607 solution video

video by MIP4U

close solution
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