# Problem related to asymptotes of hyperbola

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

• Comments are not for extended discussion; this conversation has been moved to chat. – Aloizio Macedo Jul 13 '19 at 0:15

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).

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.

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

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

• how you take x equal to e there no mention in question about x coordinate of p – Jack Rod Jul 8 '19 at 13:45
• Does not your doubleordinate go through focus? – Aqua Jul 8 '19 at 13:46
• double ordinate passes through focus – Jack Rod Jul 8 '19 at 13:48
• So? @yuvrajsingh – Aqua Jul 8 '19 at 13:50
• 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 – Jack Rod Jul 8 '19 at 13:54