Conic Sections
     Hyperbola and Line
      Construction of tangents from a point outside the hyperbola
      Properties of the hyperbola
         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
      Hyperbola and line examples
Construction of tangents from a point outside the hyperbola
With A as center draw an arc through F2, and from F1as center, draw an arc of radius 2a.  
These arcs intersect at points S1 and S2.               
Tangents are the perpendicular bisectors of the line segments F2S1 and F2S2.                                   
Tangents can also be drawn as lines through A and the intersection points of lines through F1S1 and     F1S2, with the hyperbola.                                     
These intersections are at the same time the points of contact D1 and D2.                                          
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 = 41.  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 = -
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.
Example:  From the point A(0, -3/2) drawn are tangents to the hyperbola 4x2 - 9y2 = 36, find the equations of the tangents and the area of the triangle which both tangents form with asymptotes.
Solution:  Axes of the hyperbola we read from the standard form of equation,
  4x2 - 9y2 = 36 | 36
We find tangents by solving the system of               equations,
       (1)  y = mx + c   <=  A(0, -3/2)
       (2)  a2m2 - b2 = c2   <=  (1)  c = -3/2
                                     
9m2 - 4 = (-3/2)2,   9m2 = 25/4m1,2 = 5/6
thus, the equations of the tangents,
t1 ::   y = 5/6x - 3/2  or  5x - 6y - 9 = 0  and   t ::   y = -5/6x - 3/2  or  5x + 6y + 9 = 0.
The area of the triangle that tangents form with asymptotes we calculate using the formula,
AD= (x2y1 - x1y2)/2, where S1(x1, y1) and S2(x2, y2) are the intersections (third vertex is the origin (0, 0)).
By solving system of equations,
Therefore, the intersections S1(1, -2/3) and S2(9, 6).
Then, the area of the triangle AD= (x2y1 - x1y2) / 2  gives  AD= [16 - 9(-2/3)] / 2 = 6 square units.
We can get the same result using the property that the area of the triangle which the tangent form
with asymptotes of the hyperbola is of the constant value
A = a b, so that A = 3 2 = 6.
Conic sections contents
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