HyperbolaProperty2

Hyperbola property II

Given two intersecting lines e = OB, f = OC and a point D not on these lines, consider all lines through D and their intersection points E, F with e, f. Show that the locus of the point X dividing segment EF in a given ratio r is a hyperbola with asymptotics parallel to lines {OB,OC}.

It is natural to consider the coordinate system with axes the lines {OB,OC}.
Then calculate first the coefficients of line DX and its intersections with these axes.
The coefficients of the line EF are given by the vector product (x,y,1) x (a,b,1) = (y-b, a-x, bx-ay).
Its intersections with the axes are
E(0,y'), with y' = (ay-bx)/(a-x), and
F(x',0), with x' = (ay-bx)/(y-b).
The basic relation results from the similar triangles:
y/(y-y') = r <=> y = r(y-y') <=> y = r(y - (ay-bx)/(a-x)) <=> y = r(bx-xy)/(a-x) <=>
xy = Ux + Vy, with U = rb/(r-1) and V = a/(1-r).
Translate to the coordinate system (x*,y*) such that
x = x* + d₁, y = y* + d₂ , by which the equation becomes
xy = x*y* + x*d₂ + y*d₁ + d₁d₂ = Ux+Vy = U(x*+d₁)+V(y*+d₂) = Ux* + Vy* + (Ud₁+Vd₂).
Selecting d₁ = V and d₂ = U, we get the new equation
x*y* = UV.
This short calculation proves the claim and delivers also the basis of a precise drawing of the hyperbola. The results of HyperbolaProperty.html (*) can be easily deduced from these here. In fact, it is easy now to show that the locus of the symmetric of O with respect to X is also a hyperbola. The problem in (*) corresponds then to the special case r = -1, making X the middle of EF.

Hyperbola property II

See Also