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?

  • $\begingroup$ In your Attempt, you appear to be assuming that $P$ is the point of tangency, but this is not stated in the description of the tangent. Also, you are assuming that the other hyperbola is centered at the origin, with axes aligned with the coordinate axes, though is also not stated in the description. Please update the problem statement, or otherwise clarify these issues. $\endgroup$ – Blue Feb 26 '17 at 6:15
  • $\begingroup$ @Blue Edited the question to make it clear that P IS the point of tangency. About the centre of the conic, I am not so sure. I can only guess that since this is a general question, assuming an origin centered conic would make sense. $\endgroup$ – Shashank Holla Feb 26 '17 at 11:11
  • $\begingroup$ Can you tell us the source of the problem? $\endgroup$ – Navin Feb 26 '17 at 12:12
  • $\begingroup$ I think that the problem is wrongly stated.In fact there are infinite number of such hyperbolas possible and thus infinite sum of eccentricities.The proper statement should have been sum of distinct values of eccentricities of the hyperbola. $\endgroup$ – Navin Feb 26 '17 at 12:45
  • $\begingroup$ @navinstudent Source of this problem is from a practice paper for JEE Main. About distinct eccentricities, Yes probably that would make sens $\endgroup$ – Shashank Holla Feb 26 '17 at 15:01

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.

enter image description here

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

  • $\begingroup$ The graph is beautiful, so is the solution! Thanks $\endgroup$ – Shashank Holla Feb 26 '17 at 11:13

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?

  • $\begingroup$ Using your method, I get e to be +/- 7/3. Therefore, the sum of possible values of e become 0. But @Blue gives e to be 5/3 and 5/4 $\endgroup$ – Shashank Holla Feb 27 '17 at 5:53

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