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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

Sarthak Rout
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  • The orthoptic for an ellipse and a hyperbola is its director circle. Please google 'director circle' . – Sarthak Rout Dec 07 '18 at 09:10
  • Google “orthoptic” and “director circle” yourself. The orthoptic of an ellipse is its director circle, but the orthoptic of a hyperbola is neither of its director circles. – amd Dec 07 '18 at 09:20
  • Look, I googled and I got this Wiki: In geometry, the director circle of an ellipse or hyperbola is a circle consisting of all points where two perpendicular tangent lines to the ellipse or hyperbola cross each other. And Orthoptic of a hyperbola as per wiki" The orthoptic of a hyperbola x^2/a^2 − y^2/b^2 = 1, a > b, is the circle x^2 + y^2 = a^2 − b^2 (in case of a ≤ b there are no orthogonal tangents, see below)". And , that is the director circle of the hyperbola. Please elaborate. – Sarthak Rout Dec 07 '18 at 09:29
  • There appear to be conflicting articles and terminology (not the first time). In https://en.wikipedia.org/wiki/Orthoptic_(geometry) a hyperbola’s orthoptic is pointedly not called its director circle; https://en.wikipedia.org/wiki/Hyperbola also avoids calling the orthoptic the director circle, but also uses the term “circular directrix” for the latter. Unlike the orthoptic, these circles exist for all hyperbolas, a useful trait if you’re going to use them for construction. A footnote suggests that the term “director circle” is used differently in German and English. – amd Dec 07 '18 at 10:10
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    At any rate, without further information or assumptions, I think that the best you can do here is to state an upper bound for the eccentricity. – amd Dec 07 '18 at 10:11

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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.

Intelligenti pauca
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  • I have edited the question. Please see, if you can arrive at a result. – Sarthak Rout Dec 09 '18 at 15:52
  • I don't see any relevant change in your question: without some other information you cannot find the eccentricity. I used different lines in the above example just to keep calculations easy, but you could do exactly the same reasoning with your data, and find the same result. – Intelligenti pauca Dec 09 '18 at 16:27