Hyperbola and Line
      Hyperbola and line relationships
      Polar and pole of the hyperbola
      Construction of the tangent at the point on the hyperbola
      Construction of tangents from a point outside the hyperbola
      Hyperbola and line examples
Hyperbola and line relationships
Let examine relationships between a hyperbola and a line passing through the center of the hyperbola, i.e., the origin. A line y = mx intersects the hyperbola at two points if
     the slope   |m| < b/a  but if  |m| > b/a  then,       
the line y = mx does not intersect the hyperbola at all.
The diameters of a hyperbola are straight lines passing through its center.
The asymptotes divide these two pencils of diameters into, one which intersects the curve at two points, and  the other which do not intersect.
A diameter of a conic section is a line which passes
through the midpoints of parallel chords.
Conjugate diameters of the hyperbola (or the ellipse)
are two diameters such that each bisects all chords drawn parallel to the other.            
As the equation of a hyperbola can be obtained from the equation of an ellipse by changing the sign of b2  
that is,     a2(1 - e2) = -a2(e2 - 1) = -b2
this way, we can use other formulas relating to the ellipse to obtain corresponding formulas for the hyperbola.
Therefore, when we examine conditions which determine position of a line in relation to a hyperbola that is, 
when solve the system of equations,             y = mx + c
b2x2 - a2y2 = a2b2
           then if,    a2m2 - b2 > c2   the line intersects the hyperbola at two points,
                         a2m2 - b2 = c2   the line is the tangent of the hyperbola,
                         a2m2 - b2 < c2   the line and the hyperbola do not intersect.
Polar and pole of the hyperbola
If from a point A(x0, y0), exterior to the hyperbola, drawn are tangents, then the secant line passing through 
the contact points, is the polar of the point
A. The point A is called the pole of the polar.
The equation of the polar is derived the same way as for the ellipse,
  b2x0x - a2y0y = a2b2  the equation of the polar of the point A(x0, y0).
Construction of the tangent at the point on the hyperbola
The tangent at the point P1(x1, y1) on the hyperbola is the bisector of the angle F1P1F2 subtended by focal 
radii, r1 and r2 at P1 .
The proof shown for the ellipse applied to the          hyperbola gives,                                          
See the title ' The angle between the focal radii at a point of the ellipse'.
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.
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, -3/2) and S2(9, 6).
Then, the area of the triangle AD= (x2y1 - x1y2) / 2  gives  AD= [16 - 9(-2/3)] = 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.
Pre-calculus contents H
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