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.
