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I am studying hyperbola now a days ,and came across a question ,before writing a question i have doubt is double ordinate of hyperbola passes through focus of hyperbola, my question as follows:

Let the double ordinate $PP'$ of hyperbola $x^2/4-y^2/3=1 $ is produced both side to meet asymptotes of hyperbola in $Q,Q'$. nThe product of $PQ'\cdot PQ$ is?

MY attempt -i determine the two equation of asymptotes of hyperbola that are $x/2-y/\sqrt3 =0$,and $x/2+y/\sqrt3=0$ and assume two point on hyperbola $(a\sec\theta, b\tan\theta)$ and $(a\sec\theta, -b\tan\theta)$ for $\theta$ i use dot product of asymptotes with x axis but i know this angle is for point $Q$ not for point $P$ now i don't know how to proceed further please help

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It appears that you are interpreting "$PQ'\cdot PQ$" as a vector dot product, whereas it's merely just a product of lengths. It's no wonder you're stuck.

Let's recap where you are: You have $P = (a\sec\theta, b\tan\theta)$ (note that $P'$ doesn't matter) for some arbitrary $\theta$, where $a=2$ and $b=\sqrt{3}$. You know that $Q$ and $Q'$ lie on the asymptotes and the vertical line through $P$; so, the $x$-coordinates are both $a\sec\theta$, and you can solve for the $y$-coordinates using the asymptote equations you found.

From there, the distances $PQ'$ and $PQ$ are differences in the $y$-coordinates, and $PQ'\cdot PQ$ is the product of those distances. I'll leave those calculations to you.


I find myself compelled to mention that this problem is much cleaner if you ignore many of the specifics.

Consider the hyperbola $$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \tag{1}$$ and let the vertical line through $P = (m,p)$ meet the asymptotes at $Q=(m,q)$ and $Q'=(m,-q)$ (where we may assume $q$ is non-negative).

enter image description here

We seek this product: $$|PQ'|\cdot|PQ| = (p-(-q))(q-p) = q^2-p^2 \tag{2}$$ To get it, let's write $p$ and $q$ in terms of $m$. Since $P$ is on the hyperbola, we have $$\frac{m^2}{a^2}-\frac{p^2}{b^2}=1 \quad\to\quad p^2= \frac{b^2}{a^2}\left(m^2-a^2\right) \tag{3}$$ Since $Q$ is on the asymptote with slope $b/a$, we have $$\frac{q}{m} = \frac{b}{a} \quad\to\quad q = \frac{b}{a}m \tag{4}$$ Therefore, the target product $(2)$ is

$$q^2 - p^2 = \frac{b^2}{a^2}m^2 - \frac{b^2}{a^2}\left(m^2-a^2\right) = b^2 \tag{$\star$}$$

Done! No hyperbola parameterization or asymptote equations needed, and we can ignore unnecessarily-messy calculations involving $\sqrt{3}$. $\square$

Note: Since $m$ cancelled, the product $(\star)$ is independent of the horizontal position of $P$. The "double-ordinate" (a term I have learned just today!) need not pass through the focus.

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  • $\begingroup$ brilliant sir thanks . $\endgroup$ – yuvraj singh Jul 8 at 13:45
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Double ordinate through focus cuts asymptote $y={\sqrt{3}\over 2}x$ at $Q$ at $x=e=\sqrt{7}$ so $y= {\sqrt{21}\over 2}$, so $Q=(\sqrt{7},{\sqrt{21}\over 2} )$ and $Q'=(\sqrt{7},-{\sqrt{21}\over 2} )$, while $ P= (\sqrt{7},{3\over 2})$, so

$$PQ \cdot PQ'= ({\sqrt{21}\over 2} -{3\over 2})({\sqrt{21}\over 2} +{3\over 2}) = 3$$

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  • $\begingroup$ how you take x equal to e there no mention in question about x coordinate of p $\endgroup$ – yuvraj singh Jul 8 at 13:45
  • $\begingroup$ Does not your doubleordinate go through focus? $\endgroup$ – Aqua Jul 8 at 13:46
  • $\begingroup$ double ordinate passes through focus $\endgroup$ – yuvraj singh Jul 8 at 13:48
  • $\begingroup$ So? @yuvrajsingh $\endgroup$ – Aqua Jul 8 at 13:50
  • $\begingroup$ sir i feel sorry for my words i am kid and talking to great mathamatician like you sir its privilege for me i hope you accept my sorry for my words ,and for focus we take x=ae but you have taken x=e $\endgroup$ – yuvraj singh Jul 8 at 13:54

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