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Hyperbola Equations and the Focal Property

Set of points in a plane whose distances to two foci have the same difference.

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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\begin{align*}P\end{align*} on the hyperbola, \begin{align*}|d_2 - d_1| = C\end{align*} for some constant \begin{align*}C\end{align*} .

What is the general form for a hyperbola that opens upwards and downwards and whose foci lie on the \begin{align*}y-\end{align*} axis? ___________________________

What is the shifted equation for a hyperbola that is centered around the point \begin{align*}(h,k)\end{align*} opening up and down? ___________________________    

Left and right? ___________________________

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Sketch the Hyperbolas:

  1. \begin{align*}\frac{y^2}{4} - \frac{(x - 1)^2}{4} = 9\end{align*}
  2. \begin{align*}\frac{(x - 2)^2}{9} - \frac{(y + 4)^2}{4} = 1\end{align*}
  3. \begin{align*}\frac{(x - 3)^2}{9} - \frac{(y + 1)^2}{16} = 1\end{align*}


Identify the equation of the hyperbola using the image:

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Asymtotes

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

_______________________    _______________________

What about the form \begin{align*}\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\end{align*}?

_______________________    _______________________

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Find the equations of the asymptotes:

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

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