Conic Sections
     Hyperbola
         Examples of hyperbola
      Equilateral or rectangular hyperbola with the coordinate axes as its asymptote
         Translation of equilateral or rectangular hyperbola with the coordinate axes as its asymptote
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(-13, 0) and F2(13, 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:  Find the angle subtended by the focal radii r1 and r2 at a point A(8, y > 0) of the hyperbola 
9x2 - 16y2 = 144.
Solution:   We determine the ordinate of the point A by plugging its abscissa into equation of the hyperbola,
   x = 8   =>   9x2 - 16y2 = 144
                  9 82 - 16y2 = 144,
                               16y2 = 432    =>    y2 = 27,    
 y1,2 = 33  so that  A(8, 33).
The equations of the lines of the radii r1 and r2, we     write using the formula of a line through two               points. Since,    
r1::   AF1 and F1(-c, 0) ,  and
then
                  
Therefore, the angle between the focal radii r1 and r2 at the point A of the hyperbola, as
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 - 13, 0)  and  F2(x0 + c, 0),   F1(-4 + 13, 0).
Equations of the asymptotes of a translated hyperbola
therefore, the asymptotes of the given hyperbola,
Example:   Write the equation of the hyperbola  9x2 - 25y2 = 225 in the vertex form.
Solution:   Using parallel shifting we should place the center of the hyperbola at S(-a, 0).
Rewrite    9x2 - 25y2 = 225 | 225
therefore,  a = 5 and  b = 3, so that S(-5, 0).
Then, the translated hyperbola with the center at  S(-5, 0) has the equation
Equilateral or rectangular hyperbola with the coordinate axes as its asymptote
The graph of the reciprocal function y = 1/x or  y = k/x is a rectangular (or right) hyperbola of which asymptotes are the coordinate axes.
If k > 0 then, the function is decreasing from zero to negative infinity and from positive infinity to zero, i.e., the graph of the rectangular hyperbola opening in the first and third quadrants as is shown in the right figure.
The vertices,
Translation of equilateral or rectangular hyperbola with the coordinate axes as its asymptote
The rational function by dividing the numerator by denominator,  
can be rewritten into where,
is the constant,  are the vertical and the horizontal asymptote respectively.
Therefore, the values of the vertical and the horizontal asymptotes correspond to the coordinates of the horizontal and the vertical translation of the reciprocal function  y = k/x as is shown in the figure below.
Pre-calculus contents H
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