Conic Sections
Hyperbola and Line

The area of a triangle which the tangent at a point on the hyperbola forms with asymptotes

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 area of a triangle which the tangent at a point on the hyperbola forms with asymptotes, is of constant
value A = a · b. Vertices of the triangle are the origin O
 and the intersections A and B of the tangent to the   hyperbola at the point P0(x0, y0) with asymptotes.  Then, (1)   t1 ::   b2x0x - a2y0y = a2b2    -the tangent (2)   y = ± (b/a) · x                        -asymptotes the solution of the system of the equations (1) and (2) gives intersections A and B. Plugging (2) into (1)
Since one vertex of the triangle is the origin O(0, 0) then the formula for the area, AD= (x2y1 - x1y2)/2  or
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,
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 = 25/4,  m1,2 = ±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= [1·6 - 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.
Intermediate algebra contents