We've updated our
Privacy Policy effective December 15. Please read our updated Privacy Policy and tap

Study Guides > College Algebra

Deriving the Equation of a Hyperbola Centered at the Origin

Let (c,0)\left(-c,0\right) and (c,0)\left(c,0\right) be the foci of a hyperbola centered at the origin. The hyperbola is the set of all points (x,y)\left(x,y\right) such that the difference of the distances from (x,y)\left(x,y\right) to the foci is constant.

Figure 4
If (a,0)\left(a,0\right) is a vertex of the hyperbola, the distance from (c,0)\left(-c,0\right) to (a,0)\left(a,0\right) is a(c)=a+ca-\left(-c\right)=a+c. The distance from (c,0)\left(c,0\right) to (a,0)\left(a,0\right) is cac-a. The sum of the distances from the foci to the vertex is

(a+c)(ca)=2a\left(a+c\right)-\left(c-a\right)=2a
If (x,y)\left(x,y\right) is a point on the hyperbola, we can define the following variables:
d2=the distance from (c,0) to (x,y)d1=the distance from (c,0) to (x,y)\begin{array}{l}{d}_{2}=\text{the distance from }\left(-c,0\right)\text{ to }\left(x,y\right)\\ {d}_{1}=\text{the distance from }\left(c,0\right)\text{ to }\left(x,y\right)\end{array}
By definition of a hyperbola, d2d1{d}_{2}-{d}_{1} is constant for any point (x,y)\left(x,y\right) on the hyperbola. We know that the difference of these distances is 2a2a for the vertex (a,0)\left(a,0\right). It follows that d2d1=2a{d}_{2}-{d}_{1}=2a for any point on the hyperbola. As with the derivation of the equation of an ellipse, we will begin by applying the distance formula. The rest of the derivation is algebraic. Compare this derivation with the one from the previous section for ellipses.
 d2d1=(x(c))2+(y0)2(xc)2+(y0)2=2aDistance Formula(x+c)2+y2(xc)2+y2=2aSimplify expressions. (x+c)2+y2=2a+(xc)2+y2Move radical to opposite side. (x+c)2+y2=(2a+(xc)2+y2)2Square both sides. x2+2cx+c2+y2=4a2+4a(xc)2+y2+(xc)2+y2Expand the squares. x2+2cx+c2+y2=4a2+4a(xc)2+y2+x22cx+c2+y2Expand remaining square. 2cx=4a2+4a(xc)2+y22cxCombine like terms. 4cx4a2=4a(xc)2+y2Isolate the radical. cxa2=a(xc)2+y2Divide by 4. (cxa2)2=a2[(xc)2+y2]2Square both sides. c2x22a2cx+a4=a2(x22cx+c2+y2)Expand the squares. c2x22a2cx+a4=a2x22a2cx+a2c2+a2y2Distribute a2. a4+c2x2=a2x2+a2c2+a2y2Combine like terms. c2x2a2x2a2y2=a2c2a4Rearrange terms. x2(c2a2)a2y2=a2(c2a2)Factor common terms. x2b2a2y2=a2b2Set b2=c2a2. x2b2a2b2a2y2a2b2=a2b2a2b2Divide both sides by a2b2 x2a2y2b2=1\begin{array}{ll}\text{ }{d}_{2}-{d}_{1}=\sqrt{{\left(x-\left(-c\right)\right)}^{2}+{\left(y - 0\right)}^{2}}-\sqrt{{\left(x-c\right)}^{2}+{\left(y - 0\right)}^{2}}=2a\hfill & \text{Distance Formula}\hfill \\ \sqrt{{\left(x+c\right)}^{2}+{y}^{2}}-\sqrt{{\left(x-c\right)}^{2}+{y}^{2}}=2a\hfill & \text{Simplify expressions}\text{.}\hfill \\ \text{ }\sqrt{{\left(x+c\right)}^{2}+{y}^{2}}=2a+\sqrt{{\left(x-c\right)}^{2}+{y}^{2}}\hfill & \text{Move radical to opposite side}\text{.}\hfill \\ \text{ }{\left(x+c\right)}^{2}+{y}^{2}={\left(2a+\sqrt{{\left(x-c\right)}^{2}+{y}^{2}}\right)}^{2}\hfill & \text{Square both sides}\text{.}\hfill \\ \text{ }{x}^{2}+2cx+{c}^{2}+{y}^{2}=4{a}^{2}+4a\sqrt{{\left(x-c\right)}^{2}+{y}^{2}}+{\left(x-c\right)}^{2}+{y}^{2}\hfill & \text{Expand the squares}\text{.}\hfill \\ \text{ }{x}^{2}+2cx+{c}^{2}+{y}^{2}=4{a}^{2}+4a\sqrt{{\left(x-c\right)}^{2}+{y}^{2}}+{x}^{2}-2cx+{c}^{2}+{y}^{2}\hfill & \text{Expand remaining square}\text{.}\hfill \\ \text{ }2cx=4{a}^{2}+4a\sqrt{{\left(x-c\right)}^{2}+{y}^{2}}-2cx\hfill & \text{Combine like terms}\text{.}\hfill \\ \text{ }4cx - 4{a}^{2}=4a\sqrt{{\left(x-c\right)}^{2}+{y}^{2}}\hfill & \text{Isolate the radical}\text{.}\hfill \\ \text{ }cx-{a}^{2}=a\sqrt{{\left(x-c\right)}^{2}+{y}^{2}}\hfill & \text{Divide by 4}\text{.}\hfill \\ \text{ }{\left(cx-{a}^{2}\right)}^{2}={a}^{2}{\left[\sqrt{{\left(x-c\right)}^{2}+{y}^{2}}\right]}^{2}\hfill & \text{Square both sides}\text{.}\hfill \\ \text{ }{c}^{2}{x}^{2}-2{a}^{2}cx+{a}^{4}={a}^{2}\left({x}^{2}-2cx+{c}^{2}+{y}^{2}\right)\hfill & \text{Expand the squares}\text{.}\hfill \\ \text{ }{c}^{2}{x}^{2}-2{a}^{2}cx+{a}^{4}={a}^{2}{x}^{2}-2{a}^{2}cx+{a}^{2}{c}^{2}+{a}^{2}{y}^{2}\hfill & \text{Distribute }{a}^{2}\text{.}\hfill \\ \text{ }{a}^{4}+{c}^{2}{x}^{2}={a}^{2}{x}^{2}+{a}^{2}{c}^{2}+{a}^{2}{y}^{2}\hfill & \text{Combine like terms}\text{.}\hfill \\ \text{ }{c}^{2}{x}^{2}-{a}^{2}{x}^{2}-{a}^{2}{y}^{2}={a}^{2}{c}^{2}-{a}^{4}\hfill & \text{Rearrange terms}\text{.}\hfill \\ \text{ }{x}^{2}\left({c}^{2}-{a}^{2}\right)-{a}^{2}{y}^{2}={a}^{2}\left({c}^{2}-{a}^{2}\right)\hfill & \text{Factor common terms}\text{.}\hfill \\ \text{ }{x}^{2}{b}^{2}-{a}^{2}{y}^{2}={a}^{2}{b}^{2}\hfill & \text{Set }{b}^{2}={c}^{2}-{a}^{2}.\hfill \\ \text{ }\frac{{x}^{2}{b}^{2}}{{a}^{2}{b}^{2}}-\frac{{a}^{2}{y}^{2}}{{a}^{2}{b}^{2}}=\frac{{a}^{2}{b}^{2}}{{a}^{2}{b}^{2}}\hfill & \text{Divide both sides by }{a}^{2}{b}^{2}\hfill \\ \text{ }\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}=1\hfill & \hfill \end{array}
This equation defines a hyperbola centered at the origin with vertices (±a,0)\left(\pm a,0\right) and co-vertices (0±b)\left(0\pm b\right).

A General Note: Standard Forms of the Equation of a Hyperbola with Center (0,0)

The standard form of the equation of a hyperbola with center (0,0)\left(0,0\right) and transverse axis on the x-axis is
x2a2y2b2=1\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}=1
where
  • the length of the transverse axis is 2a2a
  • the coordinates of the vertices are (±a,0)\left(\pm a,0\right)
  • the length of the conjugate axis is 2b2b
  • the coordinates of the co-vertices are (0,±b)\left(0,\pm b\right)
  • the distance between the foci is 2c2c, where c2=a2+b2{c}^{2}={a}^{2}+{b}^{2}
  • the coordinates of the foci are (±c,0)\left(\pm c,0\right)
  • the equations of the asymptotes are y=±baxy=\pm \frac{b}{a}x
The standard form of the equation of a hyperbola with center (0,0)\left(0,0\right) and transverse axis on the y-axis is
y2a2x2b2=1\frac{{y}^{2}}{{a}^{2}}-\frac{{x}^{2}}{{b}^{2}}=1
where
  • the length of the transverse axis is 2a2a
  • the coordinates of the vertices are (0,±a)\left(0,\pm a\right)
  • the length of the conjugate axis is 2b2b
  • the coordinates of the co-vertices are (±b,0)\left(\pm b,0\right)
  • the distance between the foci is 2c2c, where c2=a2+b2{c}^{2}={a}^{2}+{b}^{2}
  • the coordinates of the foci are (0,±c)\left(0,\pm c\right)
  • the equations of the asymptotes are y=±abxy=\pm \frac{a}{b}x
Note that the vertices, co-vertices, and foci are related by the equation c2=a2+b2{c}^{2}={a}^{2}+{b}^{2}. When we are given the equation of a hyperbola, we can use this relationship to identify its vertices and foci.
Figure 5. (a) Horizontal hyperbola with center (0,0)\left(0,0\right) (b) Vertical hyperbola with center (0,0)\left(0,0\right)

How To: Given the equation of a hyperbola in standard form, locate its vertices and foci.

  1. Determine whether the transverse axis lies on the x- or y-axis. Notice that a2{a}^{2} 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.

a. If the equation has the form x2a2y2b2=1\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 (±a,0)\left(\pm a,0\right), and the foci are located at (±c,0)\left(\pm c,0\right).

b. If the equation has the form y2a2x2b2=1\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 (0,±a)\left(0,\pm a\right), and the foci are located at (0,±c)\left(0,\pm c\right).

  1. Solve for aa using the equation a=a2a=\sqrt{{a}^{2}}.
  2. Solve for cc using the equation c=a2+b2c=\sqrt{{a}^{2}+{b}^{2}}.

Example 1: Locating a Hyperbola’s Vertices and Foci

Identify the vertices and foci of the hyperbola with equation y249x232=1\frac{{y}^{2}}{49}-\frac{{x}^{2}}{32}=1.

Solution

The equation has the form y2a2x2b2=1\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 x=0x=0, and solve for yy.
1=y249x2321=y24902321=y249y2=49y=±49=±7\begin{array}{l}1=\frac{{y}^{2}}{49}-\frac{{x}^{2}}{32}\hfill \\ 1=\frac{{y}^{2}}{49}-\frac{{0}^{2}}{32}\hfill \\ 1=\frac{{y}^{2}}{49}\hfill \\ {y}^{2}=49\hfill \\ y=\pm \sqrt{49}=\pm 7\hfill \end{array}
The foci are located at (0,±c)\left(0,\pm c\right). Solving for cc,
c=a2+b2=49+32=81=9c=\sqrt{{a}^{2}+{b}^{2}}=\sqrt{49+32}=\sqrt{81}=9
Therefore, the vertices are located at (0,±7)\left(0,\pm 7\right), and the foci are located at (0,9)\left(0,9\right).

Try It 1

Identify the vertices and foci of the hyperbola with equation x29y225=1\frac{{x}^{2}}{9}-\frac{{y}^{2}}{25}=1. Solution

Licenses & Attributions