Hyperbola and Line

The parallels to the asymptotes through the tangency point intersect asymptotes
The equation of the equilateral or rectangular hyperbola with the coordinate axes as its asymptotes

Properties of the hyperbola
- The tangency point bisects the line segment AB of the tangent between asymptotes.
 The abscissa of the midpoint of the segment AB, equals the abscissa of the tangency point. -  The parallels to the asymptotes through the tangency point intersect asymptotes at the points, C and D such that, OC = AC and  OD = BD .
Therefore, if given are asymptotes and the tangency point P0, we can construct the tangent by drawing the parallel to the asymptote  y = (b/a) · x through P0 to D. Mark endpoint B of segment OB taking D as the midpoint. Thus, the line segment P0B determines the tangent line.
On a similar way we could determine intersection A, of the tangent and another asymptote, using point C.
Since triangles, ODC, DP0C, DBP0 and CP0A, are congruent, it follows that the area of the parallelogram   ODP0C is equal to half of the area of the triangle OBA, i.e.,  A = (a · b) / 2.
Using this property we can derive the equation of the equilateral or rectangular hyperbola with the coordinate
axes as its asymptotes.
As the asymptotes of an equilateral hyperbola are      mutually perpendicular then the given parallelogram is
the rectangle.
And since the axes of the equilateral hyperbola are equal  that is a = b, then the area  A = a2 / 2.
Then, for every point in the new coordinate system
If we now change the coordinates into x and y, and          denote the constant by c, obtained is
the equation of the equilateral or rectangular hyperbola      with the coordinate axes as its asymptotes.
Hyperbola and line examples
Example:  Determine the semi-axis a such that the line 5x - 4y - 16 = 0 be the tangent of the hyperbola
9x2 - a2y2 = 9a2.
Solution:   Rewrite the equation          9x2 - a2y2 = 9a2 | ¸ 9a2
 and the equation of the tangent  5x - 4y - 16 = 0  or
Then, plug the slope and the intercept into tangency condition,
 Therefore, the given line is the tangent of the hyperbola
Example:  The line 13x - 15y - 25 = 0 is the tangent of a hyperbola with linear eccentricity (half the focal distance) cH = Ö41.  Write the equation of the hyperbola.
 Solution:   Rewrite the equation  13x - 15y - 25 = 0 or
 Using the linear eccentricity
and the tangency condition
 Thus, the equation of the hyperbola,
Example:  Find the normal to the hyperbola 3x2 - 4y2 = 12 which is parallel to the line  -x + y = 0.
 Solution:  Rewrite the equation of the hyperbola 3x2 - 4y2 = 12 | ¸12 The slope of the normal is equal to the slope of the  given line, y = x  =>   m = 1,   mt = -1/mn,  so  mt = -1 applying the tangency condition a2m2 - b2 = c2  <= mt = -1, a2 = 4 and b2 = 3 4·(-1)2 - 3 = c2   =>  c1,2 = ±1 tangents, t1 ::  y = -x + 1 and  t2 ::  y = -x - 1. The points of tangency,
The equations of the normals,
D1(4, -3) and  m = 1   =>  y - y1 = m ·(x -x1),       y + 3 = 1·(x - 4)  or   n1 ::   y = x - 7,
D2(-4, 3) and  m = 1  =>  y - y1 = m ·(x -x1),         y - 3 = 1·(x + 4)   or   n2 ::   y = x + 7,
Pre-calculus contents H