12.2 The hyperbola (Page 2/13)

Page 2 / 13

Deriving the equation of an ellipse 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. See [link] .

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) = 2 a

If $(x, y)$ is a point on the hyperbola, we can define the following variables:

\begin{array}{l} d_{2} = the distance from (- c, 0) to (x, y) \\ d_{1} = the distance from (c, 0) to (x, y) \end{array}

By definition of a hyperbola, $d_{2} - d_{1}$ is constant for any point $(x, y)$ on the hyperbola. We know that the difference of these distances is $2 a$ for the vertex $(a, 0) .$ It follows that $d_{2} - d_{1} = 2 a$ 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.

\begin{array}{l} d_{2} - d_{1} = \sqrt{{(x - (- c))}^{2} + {(y - 0)}^{2}} - \sqrt{{(x - c)}^{2} + {(y - 0)}^{2}} = 2 a & Distance Formula \\ \sqrt{{(x + c)}^{2} + y^{2}} - \sqrt{{(x - c)}^{2} + y^{2}} = 2 a & Simplify expressions . \\ \sqrt{{(x + c)}^{2} + y^{2}} = 2 a + \sqrt{{(x - c)}^{2} + y^{2}} & Move radical to opposite side . \\ {(x + c)}^{2} + y^{2} = {(2 a + \sqrt{{(x - c)}^{2} + y^{2}})}^{2} & Square both sides . \\ x^{2} + 2 c x + c^{2} + y^{2} = 4 a^{2} + 4 a \sqrt{{(x - c)}^{2} + y^{2}} + {(x - c)}^{2} + y^{2} & Expand the squares . \\ x^{2} + 2 c x + c^{2} + y^{2} = 4 a^{2} + 4 a \sqrt{{(x - c)}^{2} + y^{2}} + x^{2} - 2 c x + c^{2} + y^{2} & Expand remaining square . \\ 2 c x = 4 a^{2} + 4 a \sqrt{{(x - c)}^{2} + y^{2}} - 2 c x & Combine like terms . \\ 4 c x - 4 a^{2} = 4 a \sqrt{{(x - c)}^{2} + y^{2}} & Isolate the radical . \\ c x - a^{2} = a \sqrt{{(x - c)}^{2} + y^{2}} & Divide by 4 . \\ {(c x - a^{2})}^{2} = a^{2} {[\sqrt{{(x - c)}^{2} + y^{2}}]}^{2} & Square both sides . \\ c^{2} x^{2} - 2 a^{2} c x + a^{4} = a^{2} (x^{2} - 2 c x + c^{2} + y^{2}) & Expand the squares . \\ c^{2} x^{2} - 2 a^{2} c x + a^{4} = a^{2} x^{2} - 2 a^{2} c x + a^{2} c^{2} + a^{2} y^{2} & Distribute a^{2} . \\ a^{4} + c^{2} x^{2} = a^{2} x^{2} + a^{2} c^{2} + a^{2} y^{2} & Combine like terms . \\ c^{2} x^{2} - a^{2} x^{2} - a^{2} y^{2} = a^{2} c^{2} - a^{4} & Rearrange terms . \\ x^{2} (c^{2} - a^{2}) - a^{2} y^{2} = a^{2} (c^{2} - a^{2}) & Factor common terms . \\ x^{2} b^{2} - a^{2} y^{2} = a^{2} b^{2} & Set b^{2} = c^{2} - a^{2} . \\ \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}} & Divide both sides by a^{2} b^{2} \\ \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 \end{array}

This equation defines a hyperbola centered at the origin with vertices $(\pm a, 0)$ and co-vertices $(0 \pm b) .$

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

\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1

where

the length of the transverse axis is $2 a$
the coordinates of the vertices are $(\pm a, 0)$
the length of the conjugate axis is $2 b$
the coordinates of the co-vertices are $(0, \pm b)$
the distance between the foci is $2 c,$ where $c^{2} = a^{2} + b^{2}$
the coordinates of the foci are $(\pm c, 0)$
the equations of the asymptotes are $y = \pm \frac{b}{a} x$

See [link] a .

The standard form of the equation of a hyperbola with center $(0, 0)$ and transverse axis on the y -axis is

\frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1

where

the length of the transverse axis is $2 a$
the coordinates of the vertices are $(0, \pm a)$
the length of the conjugate axis is $2 b$
the coordinates of the co-vertices are $(\pm b, 0)$
the distance between the foci is $2 c,$ where $c^{2} = a^{2} + b^{2}$
the coordinates of the foci are $(0, \pm c)$
the equations of the asymptotes are $y = \pm \frac{a}{b} x$

See [link] b .

Note that the vertices, co-vertices, and foci are related by the equation $c^{2} = a^{2} + b^{2} .$ When we are given the equation of a hyperbola, we can use this relationship to identify its vertices and foci.

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Source: OpenStax, Algebra and trigonometry. OpenStax CNX. Nov 14, 2016 Download for free at https://legacy.cnx.org/content/col11758/1.6

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