Conic Sections
Hyperbola and Line The equation of the tangent at the point on the hyperbola   The equation of the tangent at the point on the hyperbola
As we already mentioned, the points of contact of a line and the hyperbola can be obtained from the
corresponding formula for the ellipse by changing
b2 with -b2 thus the tangency point or the point of contact.
 So, the intercept and slope of the tangent or b2x1x - a2y1y = a2b2 the equation of the tangent at a point P1(x1, y1) on the hyperbola.
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, or See the title ' The angle between the focal radii at a point of the ellipse'. 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 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°.   Conic sections contents 