Hyperbola and Line Hyperbola and line relationships The equation of the tangent at the point on the hyperbola   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.
Condition for a line to be the tangent to the hyperbola - tangency condition
 A line is the tangent to the hyperbola if a2m2 - b2 = c2.
Regarding the asymptotes, to which c = 0, this condition gives |m| = b/a and that is way we can say that the hyperbola touches the asymptotes at infinity.
From the tangency condition it also follows that the slopes of the tangents will satisfy the condition, That is, the tangents to the hyperbola can only be parallel to a line belonging to the pencil of lines that do not intersects 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.
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'. 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 = 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,   Pre-calculus contents H 