Hyperbolas Not Centered at the Origin
Like the graphs for other equations, the graph of a hyperbola can be translated. If a hyperbola is translated [latex]h[/latex] units horizontally and [latex]k[/latex] units vertically, the center of the hyperbola will be [latex]\left(h,k\right)[/latex]. This translation results in the standard form of the equation we saw previously, with [latex]x[/latex] replaced by [latex]\left(x-h\right)[/latex] and [latex]y[/latex] replaced by [latex]\left(y-k\right)[/latex].
hyperbolas centered at (h,k)
The standard form of the equation of a hyperbola with center [latex]\left(h,k\right)[/latex] and transverse axis parallel to the x-axis is
[latex]\frac{{\left(x-h\right)}^{2}}{{a}^{2}}-\frac{{\left(y-k\right)}^{2}}{{b}^{2}}=1[/latex]
where
- the length of the transverse axis is [latex]2a[/latex]
- the coordinates of the vertices are [latex]\left(h\pm a,k\right)[/latex]
- the length of the conjugate axis is [latex]2b[/latex]
- the coordinates of the co-vertices are [latex]\left(h,k\pm b\right)[/latex]
- the distance between the foci is [latex]2c[/latex], where [latex]{c}^{2}={a}^{2}+{b}^{2}[/latex]
- the coordinates of the foci are [latex]\left(h\pm c,k\right)[/latex]
The asymptotes of the hyperbola coincide with the diagonals of the central rectangle. The length of the rectangle is [latex]2a[/latex] and its width is [latex]2b[/latex]. The slopes of the diagonals are [latex]\pm \frac{b}{a}[/latex], and each diagonal passes through the center [latex]\left(h,k\right)[/latex]. Using the point-slope formula, it is simple to show that the equations of the asymptotes are [latex]y=\pm \frac{b}{a}\left(x-h\right)+k[/latex].
The standard form of the equation of a hyperbola with center [latex]\left(h,k\right)[/latex] and transverse axis parallel to the y-axis is
[latex]\frac{{\left(y-k\right)}^{2}}{{a}^{2}}-\frac{{\left(x-h\right)}^{2}}{{b}^{2}}=1[/latex]
where
- the length of the transverse axis is [latex]2a[/latex]
- the coordinates of the vertices are [latex]\left(h,k\pm a\right)[/latex]
- the length of the conjugate axis is [latex]2b[/latex]
- the coordinates of the co-vertices are [latex]\left(h\pm b,k\right)[/latex]
- the distance between the foci is [latex]2c[/latex], where [latex]{c}^{2}={a}^{2}+{b}^{2}[/latex]
- the coordinates of the foci are [latex]\left(h,k\pm c\right)[/latex]
Using the reasoning above, the equations of the asymptotes are [latex]y=\pm \frac{a}{b}\left(x-h\right)+k[/latex].
Like hyperbolas centered at the origin, hyperbolas centered at a point [latex]\left(h,k\right)[/latex] have vertices, co-vertices, and foci that are related by the equation [latex]{c}^{2}={a}^{2}+{b}^{2}[/latex]. 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.
- Determine whether the transverse axis is parallel to the x– or y-axis.
- If the y-coordinates of the given vertices and foci are the same, then the transverse axis is parallel to the x-axis. Use the standard form [latex]\frac{{\left(x-h\right)}^{2}}{{a}^{2}}-\frac{{\left(y-k\right)}^{2}}{{b}^{2}}=1[/latex].
- If the x-coordinates of the given vertices and foci are the same, then the transverse axis is parallel to the y-axis. Use the standard form [latex]\frac{{\left(y-k\right)}^{2}}{{a}^{2}}-\frac{{\left(x-h\right)}^{2}}{{b}^{2}}=1[/latex].
- Identify the center of the hyperbola, [latex]\left(h,k\right)[/latex], using the midpoint formula and the given coordinates for the vertices.
- Find [latex]{a}^{2}[/latex] by solving for the length of the transverse axis, [latex]2a[/latex] , which is the distance between the given vertices.
- Find [latex]{c}^{2}[/latex] using [latex]h[/latex] and [latex]k[/latex] found in Step 2 along with the given coordinates for the foci.
- Solve for [latex]{b}^{2}[/latex] using the equation [latex]{b}^{2}={c}^{2}-{a}^{2}[/latex].
- Substitute the values for [latex]h,k,{a}^{2}[/latex], and [latex]{b}^{2}[/latex] into the standard form of the equation determined in Step 1.