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An ellipse has equation $\frac{x^2}{25}+\frac{y^2}{9} = 1$ and $P(p,q)$ is a point on the ellipse. Points $F_1$ and $F_2$ have coordinates $(-4,0)$ and $(4,0)$. Show that the sum of the distances $|PF_1|$ +$|PF_2|$ does not depend on the value of p.

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First, to find distance $|F_1P|$

$|F_1P| = \sqrt{(p+4)^2 + q^2}$

as $\frac{p^2}{25}+\frac{q^2}{9} = 1$

$\frac{9p^2}{25}+q^2 = 9$

$q^2 = 9-\frac{9p^2}{25}$

$|F_1P| = \sqrt{p^2+8p+16 + 9-\frac{9p^2}{25}}$

$= \sqrt{\frac{16}{25}p^2+8p-25}$

$= \sqrt{\frac{1}{25}(16p^2+200p-625)}$

$= \frac{1}{5}\sqrt{(4p+25)^2}$

$= \frac{1}{5}(4p+25)$

Finding the distance $|F_2P|$

$|F_2P| = \sqrt{(4-p)^2 + q^2}$

$= \sqrt{p^2-8p+16 + 9-\frac{9p^2}{25}}$

$= \sqrt{\frac{16}{25}p^2-8p-25}$

$= \sqrt{\frac{1}{25}(16p^2-200p-625)}$

$= \frac{1}{5}\sqrt{(4p-25)^2}$

$= \frac{1}{5}(4p-25)$

Therefore

$|F_1P|+|F_2P|= \frac{1}{5}(4p-25)+\frac{1}{5}(4p+25) = \frac{8}{5}p$

This is exactly what we're trying to disprove. I must have made a mistake somewhere. Please can someone explain where I went wrong. My answer appears to be the negative of what it should be. I expect the correct answer will be 10 units but for some reason it does not seem to work.

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    $\begingroup$ Compare $4p$ with 25 and think about the value of $\sqrt{(4p-25)^2}$. $\endgroup$ – user Mar 3 '18 at 22:02
  • $\begingroup$ As a piece of generic advice, pick a value for $p$ and find the first line that's false. $\endgroup$ – Ben Millwood Mar 4 '18 at 8:26
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The key mistake you made here is your simplification of square roots. Actually, since a square root is positive always, you should get $\vert F_1P\vert = \frac{|4p+25|}{5} $ and $ \vert F_2P\vert = \frac{|4p-25|}{5}$. However, note $ -5 \leq p \leq 5$, so $ -20 \leq 4p \leq 20$ and $\frac{|4p+25|}{5} = \frac{4p+25}{5}, \frac{|4p-25|}{5} = \frac{-4p+25}{5}$ Summing the two distances, $p$ cancels out.

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Note that $y=0$ gives $x=\pm5$, so $-5 \le p \le 5$ for all points $P(p,q)$ on the ellipse. Therefore:

$$ |F_1P| = \frac{1}{5}\sqrt{(4p+25)^2} = \frac{1}{5}|4p+25| = \frac{1}{5}(25+4p) \\ |F_2P| = \frac{1}{5}\sqrt{(4p-25)^2} = \frac{1}{5}|4p-25| = \color{red}{\frac{1}{5}(25-4p)} $$

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  • $\begingroup$ how do you know the absolute value of |4p−25| is 25−4p $\endgroup$ – Inquirer Mar 3 '18 at 22:07
  • $\begingroup$ @Inquirer $-5 \le p \le 5 \iff -20 \le 4p \le 20$, so $25-4p \ge 25-20 \gt 0$. $\endgroup$ – dxiv Mar 3 '18 at 22:08
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Since $$e^2= a^2-b^2 = 25-9 =16\;\; \Longrightarrow \;\;e =4$$ so the point $F_1$ and $F_2$ are focuses of ellipse and by definiton of ellipse we have $$PF_1+PF_2 = constant = 2a =10$$

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  • $\begingroup$ Bingo. It is just the definition of the ellipse. Any proof based on the familiar analytic equation assumes the definition to start with because the equation is derived from the definition. $\endgroup$ – Oscar Lanzi Mar 4 '18 at 2:30
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    $\begingroup$ I don't think that's fair. The analytic equation and the focal-points definition are equivalent. This question is asking you to prove the equivalence, so I think this answer is begging the question. $\endgroup$ – Ben Millwood Mar 4 '18 at 8:28

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