# Hyperbola

Properties of asymptotes.

First of all, we may consider the asymptotes as two specific tangents : the contact point is at infinity.
Apart this, other general properties of tangents apply as well to asymptotes. For instance :

Sorry, your browser is not Java compliant.

Perpendicular projection Q of a focus on an asymptote lies on the principal circle.
Hence OQ = a
The product of distances from focii to an asymptote equals b²
FQ.F'Q' = b², that is FQ = F'Q' = b
The tangent in apex hhas length 2b (between the les asymptotes) :
AB = FQ

Recall :
Equation of hyperbola from its axis  x²/a² - y²/b² = 1
a = OA = semi axis, c = OF = OF', a² + b² = c²

In an analytic way, the asymptote directions of the conic section ax² + bxy + cy² + dx + ey + f = 0 are obtained by solving a + bm + cm² = 0 (in homogenous coordinates, Z=0).
And they intersect at center of the conic section.

If the coordinate system axes are the asymptotes, the hyperbola equation becomes very simple :  xy = k

When the asymptotes are perpendicular, the hyperbola is said "equilateral".

#### Secant to an hyperbola

Sorry, your browser is not Java compliant. Let a secant intersecting hyperbola in M and N, and intersecting the asymptotes in P and Q

MN and PQ have same midpoint

This can be proved easily from the hyperbola equation :
Let M and N the intersect points of line x/a + y/b = 1 and hyperbola xy = k.
The abscisses of M and N are then solutions of x²/a + k/b = x, that is x² - ax + ak/b = 0.
The sum of roots being a, midpoint of MN has abscisses a/2, which is the same as the midpoint I of PQ : P has coordinates (a,0) and Q = (0,b).

#### Tangent

Sorry, your browser is not Java compliant. Going to limit of the previous secant, we get :

The touch point is the midpoint of PQ

#### Normal

The normal is then the perpendicular bisector of the tangent restricted by the asymptotes.
Let I and J the intersections of normal with axes.

 OPQIJ are on a same circle.

Axes being the angle bisectors of asymptotes intersect the perpendicular bisector of PQ on the circumcircle to OPQ.

#### Area

If (x,y) is the touch point with the tangent (hence on hyperbola) P and Q have coordinates (2x, 0) (0,2y) and xy = k results into
OP.OP = 4k  multiplying by 1/2 sin(θ), angle of asymptotes :

Hyperbola is the envelope of secants PQ with Area(OPQ) = constant