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The question is:

$PQ$ is a chord joining the points $\phi_1$ and $\phi_2$ on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$. If $\phi_1\,+\,\phi_2 = 2\alpha$, where $\alpha$ is constant, prove that $PQ$ touches the hyperbola $\frac{x^2}{a^2}\cos^2\alpha-\frac{y^2}{b^2}=1$.

I found out the equation of the chord,
$$\frac{x(\tan\phi_1-\tan\phi_2)}{a}-\frac{y(\sec\phi_1-\sec\phi_2)}{b}+(\sec\phi_1\tan\phi_2-\sec\phi_2\tan\phi_1)=0$$

But how can I show that this line touches the required hyperbola?

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  • $\begingroup$ can you define $\phi_1$ and $\phi_2$ relatively to P and Q ? $\endgroup$ – Thomas Dec 26 '13 at 17:18
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    $\begingroup$ If $\phi_1$ and $\phi_2$ are points then their sum $\phi_1+\phi_2$ will also be a point. You say that $\alpha$ is a constant, i.e. a number. How can a point be equal to a number? For example, how can $(1,0)+(0,1)=7$? $\endgroup$ – Fly by Night Dec 26 '13 at 17:21
  • $\begingroup$ you cannot sum points. What is $\phi_1+\phi_2$? $\endgroup$ – Emanuele Paolini Dec 26 '13 at 17:57
  • $\begingroup$ ok! $\phi_1$ and $\phi_2$ are being used to represent two points on the hyperbola whose parameters are $\phi_1$ and $\phi_2$ respectively, i.e., the points $(a\sec\phi_1, b\tan\phi_1)$ and $(a\sec\phi_2, b\tan\phi_2)$. In the same way, $\alpha$ has been used to represent the point whose parameter is $\alpha$. As for $\phi_1\,+\,\phi_2 = 2\alpha$, here they are being referred to as angles. Sorry for the ambiguity. $\endgroup$ – Ris97 Dec 27 '13 at 5:30
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Hint: Solve the equation of the chord in terms of $\frac{y}{b}$. Then calculate $\frac{y^2}{b^2}$ and using trigonometric identities and the fact that $\phi_1+\phi_2=2\alpha$ you should obtain that $\frac{y^2}{b^2}=\frac{x^2}{a^2}cos^2{\alpha}-1$. This proves that the system of the two equations has a solution, namely the chord touches the hyperbola $\frac{x^2}{a^2}cos^2{\alpha}-\frac{y^2}{b^2}=1$. I guess by $\phi_i$ being a point you mean that $x=asec(\phi_i)$, $y=btan(\phi_i)$ is the point for $i=1,2$

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  • $\begingroup$ Yes that's what I mean by $\phi_1$ and $\phi_2$ $\endgroup$ – Ris97 Dec 27 '13 at 5:33

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