Tangent at any point on the hyperbola $x^2/9-y^2/16 = 1$ meets another hyperbola at $A$ and $B$. Tangent at any point (P) on the hyperbola $ \frac{x^2}{9}-\frac{y^2}{16}=1 $
 meets another hyperbola at $A$ and $B$. If $P$ is the midpoint of $AB$ for every choice of $P$, then floor( sum of all possible values of the eccentricities of this new hyperbola) is?
Attempt: Taking P to be $ (3\sec( \theta),4\tan( \theta)) $, the equation of tangent becomes $$ \frac{x\sec( \theta)}{3}- \frac{y\tan( \theta)}{4} =1 $$
Assuming the other hyperbola to be of the general form $ \frac{x^2}{a}-\frac{y^2}{b}=1 $
Now, should I substitute for x in the general equation from the tangent and equation the sum of the roots of that equation to $ 2(3 sec( \theta) $ and do the same thing for y and equation that to $2(4tan(\theta))$ because they are the mid points. But this seems to be a lot of work. Is there an easier method/correct method?
 A: 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.
A: Let the new hyperbola be $$\frac{x^2}{a}-\frac{y^2}{b}=1$$
Let the tangent at point $P(3secA,4tanA)$ meet the new hyperbola at $(x_1,y_1)$ and $(x_2,y_2)$.Since these two points lie on the hyperbola we can infact write$$\frac{x_1^2}{a}-\frac{y_1^2}{b}=1$$ and $$\frac{x_2^2}{a}-\frac{y_2^2}{b}=1$$Subtracting the two equations $$\frac{(x_1-x_2)(x_1+x_2)}{a}-\frac{(y_1-y_2)(y_1+y_2)}{b}=0$$which can be rewritten as $$\frac{(x_1-x_2)(3secA)}{a}-\frac{(y_1-y_2)(4tanA)}{b}=0$$.Now find $$\frac{y_1-y_2}{x_1-x_2}$$.This is same as slope of tangent to the original hyperbola.The equation of tangent at point $P$ on the original hyperbola can be written as $$\frac{xsecA}{3}-\frac{ytanA}{4}=1 $$Slope of the tangent is $$\frac{4secA}{3tanA}$$.Now substitute this in place of $$\frac{y_1-y_2}{x_1-x_2}$$ write $$\frac{b}{a}=e^2-1$$.Can you proceed?
