Conic Sections
Hyperbola and Line The area of a triangle which the tangent at a point on the hyperbola forms with asymptotes
The tangency point bisects the line segment of the tangent between asymptotes
The parallels to the asymptotes through the tangency point intersect asymptotes 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)   t  ::   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 - 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,   Conic sections contents 