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For $P(a\cos(\theta),b\sin(\theta))$ and $Q(a\cos(-\theta), b\sin(-\theta))$, which are extremities of the Latus rectum $x = ae$ of ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2}=1$, show that $PQ$ has length $\frac{2b^2}{a}$.

Ixam_338
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  • As you've just joined Math.SE, I've edited your question using MathJax. Next time, can you please do it on your own. (https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference). –  Jan 29 '18 at 05:40
  • See http://www.mathcaptain.com/geometry/latus-rectum.html# – lab bhattacharjee Jan 29 '18 at 05:41
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    Also just stating the problem without showing any effort on your side is frowned upon, and is more likely to attract downvotes than answers. What did you do about this problem, and where did you get stuck? –  Jan 29 '18 at 05:42
  • Sorry about the bad formatting and question, I’ll make sure to be clearer next time like you’ve suggested. I had a quick look at bhattacharjee’s link, it’s clearing things up for me now, my problem was just general understanding on the topic. Thanks for the help – Ixam_338 Jan 29 '18 at 05:49

1 Answers1

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Put $x=ae=\sqrt{a^2-b^2}$ into $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$,

$$\frac{a^2-b^2}{a^2}+\frac{y^2}{b^2}=1$$

$$y^2=\frac{b^4}{a^2}$$

$$y=\pm \frac{b^2}{a}$$

$$PQ=|y_1-y_2|=\frac{2b^2}{a}$$

The "extremities" mean the end points (of the latus rectum).

Ng Chung Tak
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