Given an asymptote to an hyperbola and that a line perpendicular to it, intersects it at a single point, we need to find its eccentricity.
Asymptote : $5x-4y+5=0$ and Tangent : $4x+5y-7=0$.
I thought that if we consider asymptote to be limiting tangent at infinity, then the point of intersection $(3/41,55/41)$ should lie on the director circle of the hyperbola as it is the locus of the perpendicular tangents. I am finding too many variables to handle here. And I suspect this question should be solved by an argument for a specific type of hyperbola (Maybe rectangular), but I could use a hint over here
The given data are not enough to fix the hyperbola eccentricity. You can better understand that with a simpler example: suppose you are given $y=0$ as an asymptote and $x=a$ as tangent, then take $y=mx$ as the second asymptote (with $m$ any non-vanishing real number). The equation of a hyperbola having those two asymptotes and tangent to line $x=a$ can be easily found to be $$y(y-mx)=-{a^2m^2\over4}.$$ The eccentricity of the above hyperbola is $$ e={\sqrt2\over m}\sqrt{1+m^2-\sqrt{1+m^2}}, $$ hence it can take any value between $1$ and $\sqrt{2}$, depending on the value of $m$.
EDIT.
With a rotation and a translation, the above example can be adapted to your data. You may check that the hyperbolas of equation $$ (5x-4y+5)\left((4-5m)y-(5+4m)x-5+{25\over4}m\right)={9\over64}m^2 $$ are all tangent to line $4x+5y-7=0$ and have line $5x-4y+5=0$ as asymptote, for any value of $m$. The eccentricities of these hyperbolas are given by the same formula as before.