Conic Sections
Hyperbola
Equation of the hyperbola
Properties of the hyperbola

Equilateral or rectangular hyperbola

Examples of hyperbola
Equation of the hyperbola
If in the direction of the axes we introduce a coordinate system so that the center of the hyperbola coincides with the origin, then coordinates of foci are
 F1(-c, 0) and F2(c, 0). For every point P(x, y) of the hyperbola, according to definition      | r1 - r2 | = 2a, using the formula for the distance of two points, after squaring and reducing By squaring again and grouping (c2 - a2) · x2 - a2y2 = a2 · (c2 - a2).
 Substituting c2 - a2  = b2 or obtained is b2x2 - a2y2 = a2b2 equation of the hyperbola,
 and after division by  a2b2 , the standard equation of the hyperbola.
Examining equation of the hyperbola
The hyperbola is determined by parameters, a and b, where a is the semi-major axis (or transverse semi-axis) and b is the semi-minor axis (or conjugate semi-axis). The intersection points of the hyperbola with coordinate axes we determine from its equation,
 by setting,      y = 0  =>    x = ± a the hyperbola has intercepts with the x-axis at vertices A1(-a, 0) and A2(a, 0). The segment A1A2 = 2a  is called the transverse axis. By setting,    x = 0  =>   y = ± b·i  there are no          intercepts with y-axis. Perpendicular to the transverse axis at the midpoint is the conjugate axis, whose length is  B1B2 = 2b. The hyperbola consists of two branches. If we solve the equation of the hyperbola for y,
it follows that the value of the square root will be real if  | x | > a. That is, if the absolute value of x is less then a, then y is an imaginary number, so that in the interval of x Î (-a, a)  there are no points of the hyperbola.
As the hyperbola is symmetric to both coordinate axes, we can examine behavior of only part of the curve
located at the first quadrant for values of
x > a.
Rewrite the above equation so that it represents only the positive values of the curve in this interval of x, so
 Since the value of x increases and tends to infinity,  x ® oo , then the term tends to zero, so that
 value of the square root tends to 1 and the equation of the hyperbola changes to
This equation shows that points of the hyperbola become closer and closer to this line for large values of x.
From the above equation of the hyperbola we see that for every value x > a, the value of the square root is
less then
1, it means that the ordinate y of every point of the hyperbole is less then the ordinate y’ of the
corresponding point of the line.
We also see that as x increases, the difference of ordinates, y and y’ becomes smaller, what means that points of the hyperbola become closer to this line.
 The lines, are the asymptotes of the hyperbola.
The asymptotes of a hyperbola coincide with the diagonals of the rectangle whose center is the center of the
curve, and whose sides are parallel and equal to the length
2a and 2b, of the axes of the curve, as shows the above figure.
 From the equation of the hyperbola for x =  ± c, obtained is
the semi-latus rectum p.
The length of p can also be calculated from the right triangle with legs p and 2c, whose hypotenuse is p + 2a, according to the definition of the hyperbola.
 The latus rectum
are the chords perpendicular to the transverse axis and passing through the foci.
The hyperbola which has for its transverse and conjugate axes the transverse and conjugate axes of another
 hyperbola, is said to be the conjugate hyperbola. So the hyperbolas, are
conjugate hyperbolas of each other.
A given hyperbola and its conjugate are constructed on the same reference rectangle. Thus, they have the common asymptotes and their foci lie on a circle.
Equilateral or rectangular hyperbola
The hyperbola whose semi-axes are equal, i.e., a = b
 has the equation x2 - y2 = a2.
Its asymptotes  y =  ± x are perpendicular and inclined to the x-axis at an angle of 45°.
Foci of the equilateral hyperbola,
F1(-Ö2 a, 0) and F2(Ö2 a, 0),
and the eccentricity  e = c/a = Ö2.
Translated hyperbola
The equation of a hyperbola translated from standard position so that its center is at S(x0, y0)  is given by
b2(x - x0)2 - a2(y - y0)2 = a2b2
 or
and after expanding and substituting constants obtained is
Ax2 + By2 + Cx + Dy + F = 0.
An equation of that form represents the hyperbola if
A · B < 0
that is, if coefficients of the square terms have different signs.
Examples of hyperbola
Example:  Given is the hyperbola  4x2 - 9y2 = 36,  determine the semi-axes, equations of the asymptotes,
coordinates of foci, the eccentricity and the semi-latus rectum.
Solution:   Put the equation in the standard form to
determine the semi-axes, thus
4x2 - 9y2 = 36 | ¸ 36
 Asymptotes,
 Applying,
coordinates of foci,  F1(-Ö13, 0) and F2(Ö13, 0).
 The eccentricity, and the semi-latus rectum,
 Example:  Write equation of a hyperbola with the focus at F2(5, 0) and whose asymptotes are,
 Solution:
 Therefore, the equation of the hyperbola,
Example:  The hyperbola is given by equation  4x2 - 9y2 + 32x + 54y - 53 = 0.
Find coordinates of the center, the foci, the eccentricity and the asymptotes of the hyperbola.
Solution:   The given hyperbola is translated in the direction of the coordinate axes so the values of translations x0 and y0 we can find by using the method of completing the square rewriting the equation in
 the standard form,
Thus,           4x2 + 32x - 9y2 + 54y - 53 = 0,
4(x2 + 8x) - 9(y2 - 6y) - 53 = 0
4[(x + 4)2 - 16] - 9[(y - 3)2 - 9] - 53 = 0
4(x + 4)2 - 9(y - 3)2 = 36 | ¸ 36
 Therefore,
it follows that   a2 = 9a = 3,   b2 = 4b = 2,  and the center of the hyperbola at  S(x0, y0)  or  S(-4, 3).
 Half the focal distance the eccentricity
and the foci,  F1(x0 - c, 0)  so  F1(-4 - Ö13, 0)  and  F2(x0 + c, 0),   F1(-4 + Ö13, 0).
Equations of the asymptotes of a translated hyperbola
 therefore, the asymptotes of the given hyperbola,
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