Deriving the Equation of a Hyperbola Centered at the Origin
Let (−c,0) and (c,0) be the foci of a hyperbola centered at the origin. The hyperbola is the set of all points (x,y) such that the difference of the distances from (x,y) to the foci is constant.
Figure 4
If (a,0) is a vertex of the hyperbola, the distance from (−c,0) to (a,0) is a−(−c)=a+c. The distance from (c,0) to (a,0) is c−a. The sum of the distances from the foci to the vertex is
(a+c)−(c−a)=2a
If (x,y) 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)
By definition of a hyperbola, d2−d1 is constant for any point (x,y) on the hyperbola. We know that the difference of these distances is 2a for the vertex (a,0). It follows that d2−d1=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.
d2−d1=(x−(−c))2+(y−0)2−(x−c)2+(y−0)2=2a(x+c)2+y2−(x−c)2+y2=2a(x+c)2+y2=2a+(x−c)2+y2(x+c)2+y2=(2a+(x−c)2+y2)2x2+2cx+c2+y2=4a2+4a(x−c)2+y2+(x−c)2+y2x2+2cx+c2+y2=4a2+4a(x−c)2+y2+x2−2cx+c2+y22cx=4a2+4a(x−c)2+y2−2cx4cx−4a2=4a(x−c)2+y2cx−a2=a(x−c)2+y2(cx−a2)2=a2[(x−c)2+y2]2c2x2−2a2cx+a4=a2(x2−2cx+c2+y2)c2x2−2a2cx+a4=a2x2−2a2cx+a2c2+a2y2a4+c2x2=a2x2+a2c2+a2y2c2x2−a2x2−a2y2=a2c2−a4x2(c2−a2)−a2y2=a2(c2−a2)x2b2−a2y2=a2b2a2b2x2b2−a2b2a2y2=a2b2a2b2a2x2−b2y2=1Distance FormulaSimplify expressions.Move radical to opposite side.Square both sides.Expand the squares.Expand remaining square.Combine like terms.Isolate the radical.Divide by 4.Square both sides.Expand the squares.Distribute a2.Combine like terms.Rearrange terms.Factor common terms.Set b2=c2−a2.Divide both sides by a2b2
This equation defines a hyperbola centered at the origin with vertices (±a,0) and co-vertices (0±b).
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) and transverse axis on the x-axis is
a2x2−b2y2=1
where
the length of the transverse axis is 2a
the coordinates of the vertices are (±a,0)
the length of the conjugate axis is 2b
the coordinates of the co-vertices are (0,±b)
the distance between the foci is 2c, where c2=a2+b2
the coordinates of the foci are (±c,0)
the equations of the asymptotes are y=±abx
The standard form of the equation of a hyperbola with center (0,0) and transverse axis on the y-axis is
a2y2−b2x2=1
where
the length of the transverse axis is 2a
the coordinates of the vertices are (0,±a)
the length of the conjugate axis is 2b
the coordinates of the co-vertices are (±b,0)
the distance between the foci is 2c, where c2=a2+b2
the coordinates of the foci are (0,±c)
the equations of the asymptotes are y=±bax
Note that the vertices, co-vertices, and foci are related by the equation c2=a2+b2. 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) (b) Vertical hyperbola with center (0,0)
How To: 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 a2 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 a2x2−b2y2=1, then the transverse axis lies on the x-axis. The vertices are located at (±a,0), and the foci are located at (±c,0).
b. If the equation has the form a2y2−b2x2=1, then the transverse axis lies on the y-axis. The vertices are located at (0,±a), and the foci are located at (0,±c).
Solve for a using the equation a=a2.
Solve for c using the equation c=a2+b2.
Example 1: Locating a Hyperbola’s Vertices and Foci
Identify the vertices and foci of the hyperbola with equation 49y2−32x2=1.
Solution
The equation has the form a2y2−b2x2=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=0, and solve for y.
1=49y2−32x21=49y2−32021=49y2y2=49y=±49=±7
The foci are located at (0,±c). Solving for c,
c=a2+b2=49+32=81=9
Therefore, the vertices are located at (0,±7), and the foci are located at (0,9).
Try It 1
Identify the vertices and foci of the hyperbola with equation 9x2−25y2=1.
Solution