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What is the standard form equation of the hyperbola that has vertices $\text{\hspace{0.17em}}\left(\pm 6,0\right)\text{\hspace{0.17em}}$ and foci $\text{\hspace{0.17em}}\left(\pm 2\sqrt{10},0\right)?$
The vertices and foci are on the x -axis. Thus, the equation for the hyperbola will have the form $\text{\hspace{0.17em}}\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}=1.$
The vertices are $\text{\hspace{0.17em}}\left(\pm 6,0\right),\text{\hspace{0.17em}}$ so $\text{\hspace{0.17em}}a=6\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}{a}^{2}=36.$
The foci are $\text{\hspace{0.17em}}\left(\pm 2\sqrt{10},0\right),\text{\hspace{0.17em}}$ so $\text{\hspace{0.17em}}c=2\sqrt{10}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}{c}^{2}=40.$
Solving for $\text{\hspace{0.17em}}{b}^{2},$ we have
Finally, we substitute $\text{\hspace{0.17em}}{a}^{2}=36\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}{b}^{2}=4\text{\hspace{0.17em}}$ into the standard form of the equation, $\text{\hspace{0.17em}}\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}=1.\text{\hspace{0.17em}}$ The equation of the hyperbola is $\text{\hspace{0.17em}}\frac{{x}^{2}}{36}-\frac{{y}^{2}}{4}=1,$ as shown in [link] .
What is the standard form equation of the hyperbola that has vertices $\text{\hspace{0.17em}}\left(\mathrm{0,}\pm 2\right)\text{\hspace{0.17em}}$ and foci $\text{\hspace{0.17em}}\left(\mathrm{0,}\pm 2\sqrt{5}\right)?$
$\frac{{y}^{2}}{4}-\frac{{x}^{2}}{16}=1$
Like the graphs for other equations, the graph of a hyperbola can be translated. If a hyperbola is translated $\text{\hspace{0.17em}}h\text{\hspace{0.17em}}$ units horizontally and $\text{\hspace{0.17em}}k\text{\hspace{0.17em}}$ units vertically, the center of the hyperbola will be $\text{\hspace{0.17em}}\left(h,k\right).\text{\hspace{0.17em}}$ This translation results in the standard form of the equation we saw previously, with $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ replaced by $\text{\hspace{0.17em}}\left(x-h\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ replaced by $\text{\hspace{0.17em}}\left(y-k\right).$
The standard form of the equation of a hyperbola with center $\text{\hspace{0.17em}}\left(h,k\right)\text{\hspace{0.17em}}$ and transverse axis parallel to the x -axis is
where
The asymptotes of the hyperbola coincide with the diagonals of the central rectangle. The length of the rectangle is $\text{\hspace{0.17em}}2a\text{\hspace{0.17em}}$ and its width is $\text{\hspace{0.17em}}2b.\text{\hspace{0.17em}}$ The slopes of the diagonals are $\text{\hspace{0.17em}}\pm \frac{b}{a},$ and each diagonal passes through the center $\text{\hspace{0.17em}}\left(h,k\right).\text{\hspace{0.17em}}$ Using the point-slope formula , it is simple to show that the equations of the asymptotes are $\text{\hspace{0.17em}}y=\pm \frac{b}{a}\left(x-h\right)+k.\text{\hspace{0.17em}}$ See [link] a
The standard form of the equation of a hyperbola with center $\text{\hspace{0.17em}}\left(h,k\right)\text{\hspace{0.17em}}$ and transverse axis parallel to the y -axis is
where
Using the reasoning above, the equations of the asymptotes are $\text{\hspace{0.17em}}y=\pm \frac{a}{b}\left(x-h\right)+k.\text{\hspace{0.17em}}$ See [link] b .
Like hyperbolas centered at the origin, hyperbolas centered at a point $\text{\hspace{0.17em}}\left(h,k\right)\text{\hspace{0.17em}}$ have vertices, co-vertices, and foci that are related by the equation $\text{\hspace{0.17em}}{c}^{2}={a}^{2}+{b}^{2}.\text{\hspace{0.17em}}$ We can use this relationship along with the midpoint and distance formulas to find the standard equation of a hyperbola when the vertices and foci are given.
Given the vertices and foci of a hyperbola centered at $\text{\hspace{0.17em}}\left(h,k\right),$ write its equation in standard form.
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