## 17Calculus - Hyperbolas in Rectangular Coordinates

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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

Figure 1 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

$$e=c/a$$

Figure 2 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 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 video

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

### 1605 video

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

### 1606 video

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

### 1607 video

video by MIP4U

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You CAN Ace Calculus

### Trig Formulas

The Unit Circle

The Unit Circle [wikipedia] Basic Trig Identities

Set 1 - basic identities

$$\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }$$

$$\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }$$

$$\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }$$

$$\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }$$

Set 2 - squared identities

$$\sin^2t + \cos^2t = 1$$

$$1 + \tan^2t = \sec^2t$$

$$1 + \cot^2t = \csc^2t$$

Set 3 - double-angle formulas

$$\sin(2t) = 2\sin(t)\cos(t)$$

$$\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }$$

Set 4 - half-angle formulas

$$\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }$$

$$\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }$$

Trig Derivatives

 $$\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }$$ $$\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }$$ $$\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }$$ $$\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }$$ $$\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }$$ $$\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }$$

Inverse Trig Derivatives

 $$\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }$$ $$\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }$$

Trig Integrals

 $$\int{\sin(x)~dx} = -\cos(x)+C$$ $$\int{\cos(x)~dx} = \sin(x)+C$$ $$\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C$$ $$\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C$$ $$\int{\sec(x)~dx} =$$ $$\ln\abs{\sec(x)+\tan(x)}+C$$ $$\int{\csc(x)~dx} =$$ $$-\ln\abs{\csc(x)+\cot(x)}+C$$

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