Question 58880

Jul 17, 2016

See below

Explanation:

Given an origin centered hyperbola

$h \left(x , y\right) = {a}^{2} {x}^{2} - {b}^{2} {y}^{2} - {a}^{2} {b}^{2} = 0$

and a line

$l \left(x , y\right) = c x + \mathrm{dy} + e = 0$

the tangency condition is that at tangency point ${p}_{0} = \left\{{x}_{0} , {y}_{0}\right\}$
both $h \left({x}_{0} , {y}_{0}\right) , l \left({x}_{0} , {y}_{0}\right)$ have the same declivity or the same normal vector. The normal vectors are

${\vec{n}}_{h} = \nabla h \left({x}_{0} , {y}_{0}\right) = \left\{{h}_{x} , {h}_{y}\right\} = \left\{2 {a}^{2} {x}_{0} , - 2 {b}^{2} {y}_{0}\right\}$ and
${\vec{n}}_{l} = \nabla l \left({x}_{0} , {y}_{0}\right) = \left\{{l}_{x} , {l}_{y}\right\} = \left\{c , d\right\}$

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 $a \mathrm{dn} e b c$, when the system has no solution, the system will have always an unique solution.