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Consider a family of straight line pairs given by $\frac{tx}{a}-\frac{y}{b}+t=0$ and $\frac{x}{a}+\frac{ty}{b}-1=0$ where $t$ is a parameter.

My goal is to show that the set of intersection points of the pairs forms an ellipse.

For solving this, I have multiplied the first equation by $t$ and added it to the other equation, like this:

$$\frac{t^2 x}{a}-\frac{ty}{b}+t^2=0$$ $$\frac{ x}{a}+\frac{ty}{b}-1=0$$

This gives me $$x=a(\frac{1-t^2}{t^2 +1})$$

How do I proceed from here? Particularly, how can I show that these are the coordinates of a point on an ellipse?

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Ris97
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  • Just general writing advice: using three "of"'s in quick succession makes the sentence hard to understand. Sometimes it pays to write a little more to make a sentence easier to digest :) That was what I tried to do with my revision. I hope this is ok... – rschwieb Oct 20 '14 at 13:50

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Similarly, $$y=b\frac{2t}{1+t^2}$$

$$\text{Use }(a^2-b^2)^2+(2ab)^2=(a^2+b^2)^2$$

$$a=1\implies (1-b^2)^2+(2b)^2=(1+b^2)^2$$

Alternatively, put $t=\tan\theta$ and use double angle formula

lab bhattacharjee
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  • Downvoter, please pinpoint the mistake. – lab bhattacharjee Dec 04 '13 at 04:28
  • I did not downvote, but anyone who downvote a question or an answer has no obligation to explain why he/she did so. Ok? – ILoveMath Dec 04 '13 at 04:36
  • @DonAnselmo, that is why I requested to help me to know my fault. – lab bhattacharjee Dec 04 '13 at 05:30
  • @labbhattacharjee please will you explain the last step. I can't understand that. – Ris97 Dec 04 '13 at 06:40
  • @Ris97, do you mean the alternative method? – lab bhattacharjee Dec 05 '13 at 05:37
  • No, the a=1 part – Ris97 Dec 05 '13 at 11:54
  • @Ris97, so we have $$\left(\frac{1-b^2}{1+b^2}\right)^2+\left(\frac{2ab}{1+b^2}\right)^2=1$$ – lab bhattacharjee Dec 05 '13 at 11:57
  • yes, but how does this prove that the point lies on an ellipse. Is it that these expressions have the same form as that of $x$ and $y$, and that squaring them gives $1$ like it gives for an ellipse? – Ris97 Dec 05 '13 at 12:05
  • @Ris97, you can use this(http://en.wikipedia.org/wiki/Rotation_of_axes#Identifying_rotated_conic_sections) – lab bhattacharjee Dec 05 '13 at 12:30
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    I suspect the downvoter may have felt that the answer loses the thread of the problem a bit. It may have been preferable to write the expressions as $$ \ \frac{x}{a} \ = \ \frac{1 - t^2}{1 + t^2} \ \ \text{and} \ \ \frac{ty}{b} \ = \ 1 \ - \ \frac{x}{a} \ \ \Rightarrow \ \frac{y}{b} \ = \ \frac{2t}{1 + t^2} , $$ from which it follows directly that $ \ \left( \frac{x}{a} \right)^2 \ + \ \left( \frac{y}{b} \right)^2 \ = \ 1 \ . $ [The coordinate variables disappear completely in the answer as given.] – colormegone Dec 28 '13 at 22:53