Conic Sections
Hyperbola and Line
Hyperbola and line relationships
Hyperbola and line examples
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.
Example:  Find the angle between the ellipse, which passes through points,  A(Ö5, 4/3) and B(1, 4Ö2/3), and the hyperbola whose asymptotes are y = ± x/ 2 and the linear eccentricity or half of the focal distance c = Ö5.
Solution:  Find the equation of the ellipse by solving the system of equations,
 thus, the equation of the ellipse
The equation of the hyperbola by solving the system of equations,
 Therefore, equation of the hyperbola
Angle between curves is the angle between tangents at intersection of the curves. By solving the system of equations of the curves we obtain the points of intersection,
(1)  4x2 + 9y2 = 36    (2)  => (1)   4(4y2 + 4) + 9y2 = 36,   25y2 = 20,   y1,2 = ±2/Ö5,   x1,2 = ±6/Ö5
(2)    x2 - 4y2 = 4   =>   x2 = 4y2 + 4
Tangent of the ellipse and the hyperbola at the intersection S1(6/Ö5, 2/Ö5),
S1(6/Ö5, 2/Ö5)    =>   te ::   b2x1x + a2y1y = a2b2,    4x + 3 = 6/Ö5    or    y = (-4/3)x + 2Ö5,
S1(6/Ö5, 2/Ö5)   =>   th ::   b2x1x - a2y1y = a2b2,    3x - 4 = 2/Ö5    or    y = (3/4)x - Ö5/2.
fulfilled is the perpendicularity condition. Therefore, the angle between curves j = 90°.
College algebra contents E