Given the equation of a hyperbola in standard form, locate its vertices and foci.
Determine whether the transverse axis lies on the
x - or
y -axis. Notice that
$\text{\hspace{0.17em}}{a}^{2}\text{\hspace{0.17em}}$ is always under the variable with the positive coefficient. So, if you set the other variable equal to zero, you can easily find the intercepts. In the case where the hyperbola is centered at the origin, the intercepts coincide with the vertices.
If the equation has the form
$\text{\hspace{0.17em}}\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}=1,$ then the transverse axis lies on the
x -axis. The vertices are located at
$\text{\hspace{0.17em}}(\pm a,0),$ and the foci are located at
$\text{\hspace{0.17em}}\left(\pm c,0\right).$
If the equation has the form
$\text{\hspace{0.17em}}\frac{{y}^{2}}{{a}^{2}}-\frac{{x}^{2}}{{b}^{2}}=1,$ then the transverse axis lies on the
y -axis. The vertices are located at
$\text{\hspace{0.17em}}(0,\pm a),$ and the foci are located at
$\text{\hspace{0.17em}}\left(\mathrm{0,}\pm c\right).$
Solve for
$\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ using the equation
$\text{\hspace{0.17em}}a=\sqrt{{a}^{2}}.$
Solve for
$\text{\hspace{0.17em}}c\text{\hspace{0.17em}}$ using the equation
$\text{\hspace{0.17em}}c=\sqrt{{a}^{2}+{b}^{2}}.$
Locating a hyperbola’s vertices and foci
Identify the vertices and foci of the
hyperbola with equation
$\text{\hspace{0.17em}}\frac{{y}^{2}}{49}-\frac{{x}^{2}}{32}=1.$
The equation has the form
$\text{\hspace{0.17em}}\frac{{y}^{2}}{{a}^{2}}-\frac{{x}^{2}}{{b}^{2}}=1,$ so the transverse axis lies on the
y -axis. The hyperbola is centered at the origin, so the vertices serve as the
y -intercepts of the graph. To find the vertices, set
$\text{\hspace{0.17em}}x=0,$ and solve for
$\text{\hspace{0.17em}}y.$
Therefore, the vertices are located at
$\text{\hspace{0.17em}}\left(\mathrm{0,}\pm 7\right),$ and the foci are located at
$\text{\hspace{0.17em}}\left(0,9\right).$
Just as with ellipses, writing the equation for a hyperbola in standard form allows us to calculate the key features: its center, vertices, co-vertices, foci, asymptotes, and the lengths and positions of the transverse and conjugate axes. Conversely, an equation for a hyperbola can be found given its key features. We begin by finding standard equations for hyperbolas centered at the origin. Then we will turn our attention to finding standard equations for hyperbolas centered at some point other than the origin.
Hyperbolas centered at the origin
Reviewing the standard forms given for hyperbolas centered at
$\text{\hspace{0.17em}}\left(0,0\right),$ we see that the vertices, co-vertices, and foci are related by the equation
$\text{\hspace{0.17em}}{c}^{2}={a}^{2}+{b}^{2}.\text{\hspace{0.17em}}$ Note that this equation can also be rewritten as
$\text{\hspace{0.17em}}{b}^{2}={c}^{2}-{a}^{2}.\text{\hspace{0.17em}}$ This relationship is used to write the equation for a hyperbola when given the coordinates of its foci and vertices.
Given the vertices and foci of a hyperbola centered at
$\text{\hspace{0.17em}}\left(0,\text{\hspace{0.17em}}\text{0}\right),$ write its equation in standard form.
Determine whether the transverse axis lies on the
x - or
y -axis.
If the given coordinates of the vertices and foci have the form
$\text{\hspace{0.17em}}\left(\pm a,0\right)\text{\hspace{0.17em}}$ and
$\text{\hspace{0.17em}}\left(\pm c,0\right),\text{\hspace{0.17em}}$ respectively, then the transverse axis is the
x -axis. Use the standard form
$\text{\hspace{0.17em}}\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}=1.$
If the given coordinates of the vertices and foci have the form
$\text{\hspace{0.17em}}\left(\mathrm{0,}\pm a\right)\text{\hspace{0.17em}}$ and
$\text{\hspace{0.17em}}\left(\mathrm{0,}\pm c\right),\text{\hspace{0.17em}}$ respectively, then the transverse axis is the
y -axis. Use the standard form
$\text{\hspace{0.17em}}\frac{{y}^{2}}{{a}^{2}}-\frac{{x}^{2}}{{b}^{2}}=1.$
Find
$\text{\hspace{0.17em}}{b}^{2}\text{\hspace{0.17em}}$ using the equation
$\text{\hspace{0.17em}}{b}^{2}={c}^{2}-{a}^{2}.$
Substitute the values for
$\text{\hspace{0.17em}}{a}^{2}\text{\hspace{0.17em}}$ and
$\text{\hspace{0.17em}}{b}^{2}\text{\hspace{0.17em}}$ into the standard form of the equation determined in Step 1.
Absolutely, for me. My problems with math started in First grade...involving a nun Sister Anastasia, bad vision, talking & getting expelled from Catholic school. When it comes to math I just can't focus and all I can hear is our family silverware banging and clanging on the pink Formica table.
Carole
find the 15th term of the geometric sequince whose first is 18 and last term of 387