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Prove that for any point (x,y) on the conic, the sum of the distances to the two foci is always twice the semi-major axis.

I know that this can be proven in general for all ellipses but the practice question specifically asks for this to be proven for $\frac{x^2}{9}$ + $\frac{y^2}{4}$ = 1. I feel like I'm really close but I've managed to math myself into a corner somehow.

Let the foci ($\sqrt{5}$, 0) and (-$\sqrt{5}$, 0) be denoted as F and F'. Let the point on the conic be denoted P(x,y). We are required to show PF + PF' = 2a. In this case, since a = 3, 2a = 6.

PF = $\sqrt{(x-\sqrt{5})^2 + y^2}$ and PF' = $\sqrt{(x+\sqrt{5})^2 + y^2}$

By rearranging the equation for the ellipse, we get y$^2$ = 4 - $\frac{4}{9}$x$^2$.

Substitute this into PF and PF' to get:

PF = $\sqrt{(x-\sqrt{5})^2 + 4 - \frac{4}{9}x^2}$ = $\sqrt{\frac{5}{9}x^2 - 2\sqrt{5}x + 9}$ = $\sqrt{(x - \frac{9\sqrt{5}}{5})^2}$ = x - $\frac{9\sqrt{5}}{5}$

PF' = $\sqrt{(x+\sqrt{5})^2 + 4 - \frac{4}{9}x^2}$ = $\sqrt{\frac{5}{9}x^2 + 2\sqrt{5}x + 9}$ = $\sqrt{(x + \frac{9\sqrt{5}}{5})^2}$ = x + $\frac{9\sqrt{5}}{5}$

Therefore PF + PF' = 2x

And then I got stuck

Anindya Prithvi
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    Do you need a proof $PF+PF^\prime$ is constant? Because the cases where $P$ is at either end of the semi-major axis, which passes through both foci, provide an easy computation of the constant. – J.G. Aug 04 '20 at 07:34
  • That would help for certain points but we are required to prove it for any point on the ellipse –  Aug 04 '20 at 07:43
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    Regarding your calculation $\sqrt{\frac{5}{9}x^2 + 2\sqrt{5}x + 9}$ = $\sqrt{(x + \frac{9\sqrt{5}}{5})^2}$ are you sure? – Hrishabh Nayal Aug 04 '20 at 07:52
  • @L.ABrudvik-Lindner So: yes, you do need a proof it's constant. – J.G. Aug 04 '20 at 08:00

3 Answers3

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An ellipse is a plane curve surrounding two foci, such that for all points on the curve, the sum of the two distances to the foci is a constant. One starts with $\sqrt{(x-c)^2+y^2}+\sqrt{(x+c)^2+y^2}=2a$ (your question) to arrive at $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, where $2a$ is any constant (which ends up being the length of the semi-major axis), $b^2=a^2-c^2$, and the foci are $(-c,0),(+c,0)$. Note that $a,b,c\in\mathbb R^+$.

Addendum

Finding the distance of either foci from a point $P(x,y)$ on the ellipse, $$PF=\sqrt{(x-c)^2+y^2}=\sqrt{(x-c)^2+b^2-\frac{b^2}{a^2}x^2}=\sqrt{\frac{c^2}{a^2}x^2-2cx+a^2}=\left\lvert\frac ca x-a\right\rvert=a-ex$$ $$PF^\prime=\sqrt{(x+c)^2+y^2}=\sqrt{(x+c)^2+b^2-\frac{b^2}{a^2}x^2}=\sqrt{\frac{c^2}{a^2}x^2+2cx+a^2}=\left\lvert\frac ca x+a\right\rvert=a+ex$$ since $x\in[-a,+a]$, where $e=\frac ca$.

Sameer Baheti
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  • I'm not quite seeing how the top bit connects to the bottom bit –  Aug 04 '20 at 08:22
  • It's the exact answer I would've given. The distance of any point (x,y) from the foci is (a-ex, a+ex), a is length of semi major axis, e is the eccentricity of the ellipse. Bring your expression for distance in this form, and then it's simple addition. @L.ABrudvik-Lindner – Aman Kumar Aug 04 '20 at 08:31
  • Got it. Thank you –  Aug 04 '20 at 08:36
  • @L.ABrudvik-Lindner If you think you have got it, perhaps you can approve an answer. – Sameer Baheti Aug 04 '20 at 09:05
  • Last question: How did you do the last step - to get a - ex. Because I keep getting ex - a (for PF). I'm assuming you use the quadratic formula –  Aug 04 '20 at 11:01
  • @L.ABrudvik-Lindner I have added a step $\sqrt{x^2}=\lvert x\rvert$. – Sameer Baheti Aug 04 '20 at 12:27
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For such a question, the realm of complex numbers helps a lot. In the complex plane, an ellipse is described by : $$|z-a| + |z-b|=c$$ where z represents each point on the ellipse, (a,b) is a tuple of complex numbers (which may or may not be real), c is an arbitrary constant $\geq |a-b|$

Also note, (a,b) are the focii of the ellipse as well and c is the major axis.

Coming on to the question, just describe the given equation in the form of complex numbers, the result it direct.

So, the given ellipse is $\frac{x^2}{9}$ + $\frac{y^2}{4}$ = 1 and the focii ($\sqrt{5}$, 0) and (-$\sqrt{5}$, 0).

Describe this as $$|z-\sqrt{5}| + |z-\sqrt{-5}|=6$$

This directly means that the sum of distance of any point $z$ on the ellipse is equal to the major axis.

Anindya Prithvi
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An ellipse is by definition the locus of the points such that the sum of their distances to the foci is constant.

From this property, the result is immediate:

$$PQ=FQ+QG.$$

enter image description here

  • No, it was because you didnt put a source where it originates from..or i mean how to see it intuitively. The image makes it a bit clear but before that it seemed like a comment (i mean no disrespect) – Anindya Prithvi Aug 04 '20 at 08:24
  • @AnindyaPrithvi: no source is needed for the very well-known construction of the ellipse nor for the addition of segments, is it ? –  Aug 04 '20 at 08:26
  • Actually, i sympathize with the OP, considering the method adopted it is clear that one might have to be reminded such a thing ...the reminder would often include a proof or an insight from my side (and i have removed the dv as well) – Anindya Prithvi Aug 04 '20 at 08:30
  • A point F such that the ratio of distance between the point E and the directrix and the point E and point F is a fixed constant e which is $\leq$ 1 – Anindya Prithvi Aug 04 '20 at 08:46
  • I hope you get my intentions as to why I said so...I just think the OP might not be familiar with the phrasing or the intuitions behind this specific definitions...be it trivial – Anindya Prithvi Aug 04 '20 at 08:51
  • am asking "An ellipse is by definition the locus of the points such that the sum of their distances to the foci is constant" ... how's one supposed to know this if the focus is defined to them using the eccentricity and directrix? – Anindya Prithvi Aug 04 '20 at 09:38
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    @AnindyaPrithvi: my gardener knows. –  Aug 04 '20 at 09:43