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If a chord joining the points $P(a\sec\alpha,a\tan\alpha)$ and $Q(a\sec\beta,a\tan\beta)$ on the hyperbola $x^2-y^2=a^2$ is a normal to it at $P$,then show that $\tan \beta=\tan\alpha(4\sec^2\alpha-1)$.


The chord joining the points $P(a\sec\alpha,a\tan\alpha)$ and $Q(a\sec\beta,a\tan\beta)$ on the hyperbola $x^2-y^2=a^2$ is given by $x\cos\frac{\alpha-\beta}{2}-y\sin\frac{\alpha+\beta}{2}=a\cos\frac{\alpha+\beta}{2}$....................$(1)$

And the normal to the hyperbola at $P(a\sec\alpha,a\tan\alpha)$ is given by
$\frac{x}{\sec\alpha}+\frac{y}{\tan\alpha}=2a$.............$(2)$

equation $(1)$ and $(2)$ are the same lines
Comparing them,gives
$\frac{\cos\frac{\alpha-\beta}{2}}{\frac{1}{\sec\alpha}}=\frac{-\sin\frac{\alpha+\beta}{2}}{\frac{1}{\tan\alpha}}=\frac{a\cos\frac{\alpha+\beta}{2}}{2a}$

$\sec\alpha=\frac{\cos\frac{\alpha+\beta}{2}}{2\cos\frac{\alpha-\beta}{2}}$ and $\tan\alpha=\frac{\cos\frac{\alpha+\beta}{2}}{-2\sin\frac{\alpha+\beta}{2}}$

I am stuck here.I do not know how to proceed further.

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The equation of the line $PQ$ is given by $$y-a\tan\alpha=\frac{a\tan\alpha-a\tan\beta}{a\sec\alpha-a\sec\beta}(x-a\sec\alpha),$$ i.e. $$\frac{\sin\alpha\cos\beta-\sin\beta\cos\alpha}{\sin\alpha-\sin\beta}x+\frac{\cos\alpha-\cos\beta}{\sin\alpha-\sin\beta}y=a\tag3$$

The $(2)$ you wrote can be written as $$\frac{\cos\alpha}{2}x+\frac{1}{2\tan\alpha}y=a\tag4$$

Comparing $(3)$ with $(4)$ gives $$\frac{\sin\alpha\cos\beta-\sin\beta\cos\alpha}{\sin\alpha-\sin\beta}=\frac{\cos\alpha}{2},$$ i.e. $$2\sin\alpha-\cos\alpha\tan\beta=\cos\alpha\sin\alpha\sec\beta\tag5$$ Squaring the both sides and using $\sec^2\beta=1+\tan^2\beta$ gives $$(\cos^2\alpha\sin^2\alpha-\cos^2\alpha)\tan^2\beta+4\sin\alpha\cos\alpha\tan\beta+\cos^2\alpha\sin^2\alpha-4\sin^2\alpha=0$$ and so $$\begin{align}&\tan\beta\\&=\frac{-2\sin\alpha\cos\alpha\pm\sqrt{4\sin^2\alpha\cos^2\alpha-(\cos^2\alpha\sin^2\alpha-\cos^2\alpha)(\cos^2\alpha\sin^2\alpha-4\sin^2\alpha)}}{\cos^2\alpha\sin^2\alpha-\cos^2\alpha}\\\\&=\frac{-2\sin\alpha\cos\alpha\pm\sqrt{\cos^2\alpha\sin^2\alpha(\cos^2\alpha-2)^2}}{\cos^2\alpha\sin^2\alpha-\cos^2\alpha}\\\\&=\frac{-2\sin\alpha\cos\alpha\pm \cos\alpha\sin\alpha(\cos^2\alpha-2)}{\cos^2\alpha\sin^2\alpha-\cos^2\alpha}\\\\&=\frac{-2\sin\alpha\cos\alpha - \cos\alpha\sin\alpha(\cos^2\alpha-2)}{\cos^2\alpha\sin^2\alpha-\cos^2\alpha},\quad \frac{-2\sin\alpha\cos\alpha + \cos\alpha\sin\alpha(\cos^2\alpha-2)}{\cos^2\alpha\sin^2\alpha-\cos^2\alpha}\\\\&=\tan\alpha,\quad \tan\alpha(4\sec^2\alpha-1)\end{align}$$

Now when $\tan\beta=\tan\alpha$, $$\begin{align}(5)&\Rightarrow \sin\alpha(\cos\beta-\cos\alpha)=0\\&\Rightarrow \sin\alpha=0\quad\text{or}\quad \cos\beta=\cos\alpha\end{align}$$

Case 1 : $\sin\alpha=0\Rightarrow \tan\alpha=\tan\beta=0\Rightarrow (\alpha,\beta)=(0,\pi),(\pi,0)\qquad (\text{$\tan\beta=\tan\alpha(4\sec^2\alpha-1)$ holds in this case})$

Case 2 : $\cos\beta=\cos\alpha\Rightarrow \sin\alpha=\sin\beta\Rightarrow \alpha=\beta\qquad (\text{this case is eliminated})$

Thus, we get that $\tan\beta=\tan\alpha(4\sec^2\alpha-1)$.

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As $(2)$ passes through $Q(a\sec\beta,a\tan\beta),$

$$\dfrac{\sec\beta}{\sec\alpha}=2-\dfrac{\tan\beta}{\tan\alpha}$$

Squaring and writing $\tan\alpha=p,\tan\beta=q,$

$$\dfrac{1+q^2}{1+p^2}=4+\dfrac{q^2}{p^2}-\dfrac{4q}p$$

$$\iff4-\dfrac{4q}p=\dfrac{1+q^2}{1+p^2}-\dfrac{q^2}{p^2}$$

$$\iff\dfrac{4(p-q)}p=\dfrac{p^2-q^2}{(1+p^2)p^2}$$

If $p=q$ i.e., $\tan\alpha=\tan\beta,(2)\implies\sec\beta=\sec\alpha\implies P=Q$

So, cancelling $p-q\ne0$ and assuming $p\ne0,$

$$\iff4=\dfrac{p+q}{(1+p^2)p}\iff q=4p^3+3p=p\{4(p^2+1)-1\}$$

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    $\begingroup$ I tried squaring and putting $\sec^2\beta=1+\tan^2\beta$,but i am not getting required answer in the end.@labbhattacharjee $\endgroup$ – Brahmagupta Mar 15 '16 at 12:17
  • $\begingroup$ @Brahmagupta, Please find the updated version. $\endgroup$ – lab bhattacharjee Apr 17 '16 at 10:47

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