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Isolate the variables on the left-hand side of the equation and the constants on the right-hand side:
Finally, divide both sides by ${a}^{2}-{c}^{2}.$ This gives the equation
We now define b so that ${b}^{2}={c}^{2}-{a}^{2}.$ This is possible because $c>a.$ Therefore the equation of the ellipse becomes
Finally, if the center of the hyperbola is moved from the origin to the point $\left(h,k\right),$ we have the following standard form of a hyperbola.
Consider the hyperbola with center $\left(h,k\right),$ a horizontal major axis, and a vertical minor axis. Then the equation of this ellipse is
and the foci are located at $\left(h\pm c,k\right),$ where ${c}^{2}={a}^{2}+{b}^{2}.$ The equations of the asymptotes are given by $y=k\pm \frac{b}{a}\left(x-h\right).$ The equations of the directrices are
If the major axis is vertical, then the equation of the hyperbola becomes
and the foci are located at $\left(h,k\pm c\right),$ where ${c}^{2}={a}^{2}+{b}^{2}.$ The equations of the asymptotes are given by $y=k\pm \frac{a}{b}\left(x-h\right).$ The equations of the directrices are
If the major axis (transverse axis) is horizontal, then the hyperbola is called horizontal, and if the major axis is vertical then the hyperbola is called vertical. The equation of a hyperbola is in general form if it is in the form $A{x}^{2}+B{y}^{2}+Cx+Dy+E=0,$ where A and B have opposite signs. In order to convert the equation from general to standard form, use the method of completing the square.
Put the equation $9{x}^{2}-16{y}^{2}+36x+32y-124=0$ into standard form and graph the resulting hyperbola. What are the equations of the asymptotes?
First add 124 to both sides of the equation:
Next group the x terms together and the y terms together, then factor out the common factors:
We need to determine the constant that, when added inside each set of parentheses, results in a perfect square. In the first set of parentheses, take half the coefficient of x and square it. This gives ${\left(\frac{4}{2}\right)}^{2}=4.$ In the second set of parentheses, take half the coefficient of y and square it. This gives ${\left(\frac{\mathrm{-2}}{2}\right)}^{2}=1.$ Add these inside each pair of parentheses. Since the first set of parentheses has a 9 in front, we are actually adding 36 to the left-hand side. Similarly, we are subtracting 16 from the second set of parentheses. Therefore the equation becomes
Next factor both sets of parentheses and divide by 144:
The equation is now in standard form. Comparing this to [link] gives $h=\mathrm{-2},$ $k=1,$ $a=4,$ and $b=3.$ This is a horizontal hyperbola with center at $\left(\mathrm{-2},1\right)$ and asymptotes given by the equations $y=1\pm \frac{3}{4}\left(x+2\right).$ The graph of this hyperbola appears in the following figure.
Put the equation $4{y}^{2}-9{x}^{2}+16y+18x-29=0$ into standard form and graph the resulting hyperbola. What are the equations of the asymptotes?
$\frac{{\left(y+2\right)}^{2}}{9}-\frac{{\left(x-1\right)}^{2}}{4}=1.$ This is a vertical hyperbola. Asymptotes
$y=\mathrm{-2}\pm \frac{3}{2}\left(x-1\right).$
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