Given an origin centered hyperbola
h(x,y)=a^2 x^2-b^2y^2-a^2b^2=0
and a line
l(x,y) = c x +dy+e=0
the tangency condition is that at tangency point p_0 = {x_0,y_0}
both h(x_0,y_0),l(x_0,y_0) have the same declivity or the same normal vector. The normal vectors are
vec n_h = grad h(x_0,y_0) = {h_x,h_y} = {2a^2 x_0,-2b^2 y_0} and
vec n_l = grad l(x_0,y_0) = {l_x,l_y} = {c,d}
so
vec n_h = lambda vec n_l
The determination of p_0 is obtained by solving
{
(2a^2x_0=lambda c),
(-2b^2y_0=lambda d),
(c x_0+d y_0+e=0)
:}
a,b,c,d,e are known and we need to solve for x_0,y_0,lambda
Observing the set of linear equations,
we can expect:
1) One solution
2) Infinite solutions
3) No solution
After the results
{(x_0 = -(b^2 c e)/(b^2 c^2 - a^2 d^2)),
(y_0 =(a^2 d e)/(b^2 c^2 - a^2 d^2)),
(lambda=(2 a^2 b^2 e)/(b^2 c^2 - a^2 d^2))
:}
we can see that unless adne bc, when the system has no solution, the system will have always an unique solution.
