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Hyperbola Equations and the Focal Property

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Conic Sections: Hyperbolas
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Vocabulary

Complete the chart.
 Word Definition Hyperbola ________________________________________________________________ ________________ a shape is so large that no circle, no matter how large, can enclose the shape ________________ a line which a curve approaches as the curve and the line approach infinity Perpendicular hyperbola ________________________________________________________________

Hyperbolas

Hyperbolas two foci, and they can be defined as the set of points in a plane whose distances to these two points have the same difference . So in the picture below, for every point$P$ on the hyperbola, $|d_2 - d_1| = C$ for some constant $C$ .

What is the general form for a hyperbola that opens upwards and downwards and whose foci lie on the $y-$ axis? ___________________________

What is the shifted equation for a hyperbola that is centered around the point $(h,k)$ opening up and down? ___________________________

Left and right? ___________________________

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Sketch the Hyperbolas:

1. $\frac{y^2}{4} - \frac{(x - 1)^2}{4} = 9$
2. $\frac{(x - 2)^2}{9} - \frac{(y + 4)^2}{4} = 1$
3. $\frac{(x - 3)^2}{9} - \frac{(y + 1)^2}{16} = 1$

Identify the equation of the hyperbola using the image:

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Asymtotes

For a hyperbola of the form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ , which lines are the asymtotes?

_______________________    _______________________

What about the form $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$?

_______________________    _______________________

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Find the equations of the asymptotes:

1. $\frac{(x - 1)^2}{1} - \frac{9(y + 4)^2}{1} = 9$
2. $\frac{(y + 2)^2}{16} - \frac{(x - 2)^2}{1} = 1$
3. $\frac{(x - 4)^2}{1} - \frac{(y + 1)^2}{4} = 1$
4. $\frac{y^2}{16} - \frac{(x + 1)^2}{4} = 1$
Graph the hyperbolas, give the equation of the asymptotes, use the asymptotes to enhance the accuracy of your graph.
1. $\frac{(x - 2)^2}{16} - \frac{(y + 4)^2}{1} = 1$
2. $\frac{(x + 2)^2}{9} - \frac{(y + 2)^2}{16} = 1$
3. $\frac{(x + 4)^2}{9} - \frac{(y - 2)^2}{4} = 1$

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