conic sectionsgraphing-functionstrigonometry
I know that the parametric equation for points on a hyperbola($\frac{x^2}{a^2}-\frac{y^2}{b^2} = 1$) is:
$$x = a\sec \theta$$ $$y = b\tan \theta$$
However, what does the parameter $\theta$ actually represent? In a circle, it is quite obvious, what the parameter $\theta$ represents. However, how should I visualize it for a hyperbola?
I hope the following plot helps you. I made it by Maple environment using below codes:
[> with(plots):
a := ln(3);
complexplot(2*exp(-a*t)*exp((1/2)*(2*I)*Pi*t+(1/2)*Pi), t = 0 .. 4);
I'll work in a bit more generality, in order to keep track of a few parameters better. Based on your Attempt, my reading of the problem goes like this:
For each $P$ on the hyperbola $$H := \quad\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \tag{1}$$ we have that $P$ is the midpoint of the segment determined by the two points where tangent line at $P$ meets a second (and third) hyperbola $$K_{k} :=\quad \frac{x^2}{m^2} - \frac{y^2}{n^2} = k \qquad \tag{2}$$ (where we can take all of $a$, $b$, $m$, $n$, to be positive, and where $k = \pm 1$). What can we say about the eccentricity of the other hyperbola(s)?
As you did, we parameterize $P$ by $$P = ( a \sec \theta, b \tan \theta )$$ A tangent vector at $P$ is then given by $$v = ( a \tan \theta, b \sec \theta )$$ Given the midpoint property of $P$, we know that the second hyperbola contains the points $$P \pm r v$$ for some $r$ (that depends upon $\theta$). Substituting these points into $(2)$ (and defining $t := \tan\theta$, $s := \sec\theta$ for notational simplicity) gives $$\frac{a^2}{m^2}( s \pm r t )^2 - \frac{b^2}{n^2}( t \pm r s )^2 = k \tag{3}$$ Subtracting the "$-$" version of $(3)$ from the "$+$" version cancels everything but the cross terms involving $2rst$, and we conclude $$\frac{a^2}{m^2}(4rst) = \frac{b^2}{m^2}(4rst) \quad\to\quad \frac{a}{b} = \frac{m}{n} \quad\text{(since all parameters are positive)}\tag{4}$$
Consequently, $K_{+}$ has the same "transverse-conjugate axis ratio" as $H$, and thus also the same eccentricity; on the other hand, $K_{-}$ has the eccentricity of the conjugate hyperbola of $H$.
$$\operatorname{ecc} K_{+} = \operatorname{ecc} H = \frac{\sqrt{a^2 + b^2}}{a} \quad\text{and}\quad \operatorname{ecc} K_{-} = \operatorname{ecc} \left(\text{conj of } H \right) = \frac{\sqrt{a^2+b^2}}{b} \tag{$\star$}$$
Here's an image, showing $H$ (in green) with two instances of $K_{+}$ (in blue) and two instances of $K_{-}$ (in red). (Note that they all have common asymptotes.) The circles about the point of tangency demonstrate that the tangent line meets the hyperbolas at equidistant points.
Finding the floor of the sum of the eccentricities when $a = 3$ and $b = 4$ is left as an exercise to the reader.
Best Answer
Here is an animated gif of $\theta$ changing.
Excuse the red frames (90 and 270 degree). That's my software hitting the discontinuity.