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 

 

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