17Calculus - Hyperbolas in Rectangular Coordinates
17Calculus
On this page, we discuss hyperbolas in rectangular coordinates. For a discussion of hyperbolas in polar form, see this separate page.
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Hyperbola
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Figure 1 |
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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.
\(\displaystyle{ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 }\) | |
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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 | \( e=c/a \) |
\(\displaystyle{ \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 }\) | |
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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 |
\( e=c/a \) |
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Figure 2 |
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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 |
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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
PatrickJMT - 1592 video solution
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
PatrickJMT - 1604 video solution
video by PatrickJMT |
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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
PatrickJMT - 1605 video solution
video by PatrickJMT |
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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
MIP4U - 1606 video solution
video by MIP4U |
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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
MIP4U - 1607 video solution
video by MIP4U |
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