Tangent at any point on the hyperbola $x^2/9-y^2/16 = 1$ meets another hyperbola at $A$ and $B$.

Question

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?

Answer 1

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.

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Finding the floor of the sum of the eccentricities when $a = 3$ and $b = 4$ is left as an exercise to the reader.

Answer 2

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?