My question is given a ellipse of the equation : $$\frac{x^2}{a^2}+\frac{y^2}{b^2} =1$$ where $a>b$ then how we can find the coordinates of the foci. I want to find those coordinates without the presuming that the foci exists because most proofs I found online assume the properties of foci to be true and then take some extreme case to find foci's coordinates.

So to rephrase my question: Given a closed curve of the equation $$\frac{x^2}{a^2}+\frac{y^2}{b^2} =1$$ prove that there exists two points inside the curve such that if we take any point on the boundary of the curve and join it to those two points then the sum of those lengths will give a fixed constant based on $a$ and $b$ (Assume $a>b$).

Here's my attempt which gave me nothing:

Let's take a point $P$ on the curve as $\left(x,\ b\sqrt{1-\frac{x^2}{a^2}}\right)$

Let those two points be $(-f,0)$ and $(f,0)$ then the sum of lengths from point $P$ becomes

$$S = \sqrt{(f+x)^2+b^2\left(1-\frac{x^2}{a^2}\right)}+\sqrt{(f-x)^2+b^2\left(1-\frac{x^2}{a^2}\right)}$$

Differentiating this wrt to $f$ and equating to $0$ to find the stationary case


Squaring and simplifying

$$4bfx\left(1-\frac{x^2}{a^2}\right)=0 \implies f =0$$

which is obviously wrong...

Note that in my attempt I too assumed two points, that the foci will be symmetrical and on the major axis, if we can even take these assumptions out that would be amazing. It's just that with these assumptions I was able to at least start somewhere.


$\textbf{thanks to the comments by @Blue}$

differentiating $S$ wrt $x$ and equating to zero

$$\dfrac{2\left(x+f\right)-\frac{2b^2x}{a^2}}{2\sqrt{\left(x+f\right)^2+b^2\cdot\left(1-\frac{x^2}{a^2}\right)}}+\dfrac{-\frac{2b^2x}{a^2}-2\left(f-x\right)}{2\sqrt{b^2\cdot\left(1-\frac{x^2}{a^2}\right)+\left(f-x\right)^2}}=0$$ Squaring and simplifying

$$ a^4b^2fx(f^2-(a^2-b^2))=0$$

So, $$f= \sqrt{a^2-b^2}$$

  • 1
    $\begingroup$ Do the following: find the eccentricity of the ellipse. Then the coordinates of foci will be $F_1(ae,0)$ and $F_2(-ae,0)$ where $a$ is your coefficient from the equation. $\endgroup$
    – Rusurano
    Feb 23, 2023 at 8:04
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    $\begingroup$ You're assuming that $f$ and $S$ are constant as $x$ changes. So, your derivative of $S$ should be with respect to $x$; the result is a little more complicated than what you have written. (Also, note that your square root expressions should have $b^2$ instead of $b$.) $\endgroup$
    – Blue
    Feb 23, 2023 at 8:15
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    $\begingroup$ Your derivative of $S$ with respect to $x$ is correct, but the conclusion that the derivative being $0$ implies $x=0$ is not. (Indeed, you don't want to conclude that $x$ has to be a specific value, since you're trying to prove that $S$ is constant for all $x$.) $\endgroup$
    – Blue
    Feb 23, 2023 at 8:38
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    $\begingroup$ @Blue then what should I do after taking the derivative, I am able to understand what you're saying but can you elaborate a little bit more $\endgroup$ Feb 23, 2023 at 9:02
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    $\begingroup$ @sparrow_2764: (Simplifying a bit) You have $$\frac{a^2(x+f)-b^2x}{\sqrt{a^2(x+f)^2+b^2(a^2-x^2)}}=\frac{b^2x+a^2(f-x)}{\sqrt{b^2(a^2-x^2)+a^2(f-x)^2}}$$ Squaring and combining gives $$a^4b^2fx(f^2-(a^2-b^2))=0$$ Since $a$ and $b$ are presumably non-zero, we ignore those factors. Since $a>b$, we know $f\neq0$ (why?), we ignore that factor, too. Thus, either $x=0$ or $f^2=a^2-b^2$. If the former were required, then there'd be no "foci" (as the constant-distance-sum property wouldn't hold for all points on the ellipse). However, you can verify that the latter yields a constant $S$. $\endgroup$
    – Blue
    Feb 23, 2023 at 9:54

2 Answers 2

  1. Find the eccentricity of the ellipse.


  1. Now you have the following expressions to find the coordinates of the foci:

$$\begin{align*} F_1 &= \left(ae, 0\right) = \left(a\sqrt{1-\dfrac{b^2}{a^2}}, 0\right),\\ F_2 &= \left(-ae, 0\right) = \left(-a\sqrt{1-\dfrac{b^2}{a^2}}, 0\right). \end{align*}$$

What's even more wonderful, it works for any kind of ellipse.

  • $\begingroup$ that's all nice and good but how did you prove that there are only two distinct points which qualify for the foci and how did you get this relation between foci and eccentricity. I am a lil bit new to coord geometry so I don't actually know much about what is eccentricity... $\endgroup$ Feb 23, 2023 at 8:11
  • $\begingroup$ So, what are you after exactly: expressing the coordinates of the foci via the coefficients of the ellipse equation, or proving that two foci exist for any ellipse? $\endgroup$
    – Rusurano
    Feb 23, 2023 at 8:16
  • $\begingroup$ first proving that they exist then finding their coords, I know my english is a bit choppy in the questions so please don't mind $\endgroup$ Feb 23, 2023 at 8:17
  • $\begingroup$ The existence of foci for any ellipse is straight from its definition mentioned here in WolframAlpha. For the distance, see my answer. If you are for some reason not satisfied, then what is your definition of the ellipse? $\endgroup$
    – Rusurano
    Feb 23, 2023 at 8:26

The sum of lengths of $(-f, 0)$ and $(f, 0)$ from point P is $$S = \sqrt{(f+x)^2+b^2\left(1-\frac{x^2}{a^2}\right)}+\sqrt{(f-x)^2+b^2\left(1-\frac{x^2}{a^2}\right)}$$ Your approach was correct upto this step. However, you need to realize what happens when you differentiate S with respect to f.

The condition $\frac{{\partial}S}{{\partial}f} = 0$ considers the points $(-f, 0)$ and $(f, 0)$ to be variable, and finds the minimum value of the sum of distances $S$. Intuitively, the sum of distances is minimum when both the points coincide with the center of the ellipse. Hence, you get the value $f = 0$.

What you really need to do to find the focal points is to find a value of $f$ such that the expression for $S$ is independent of the parameter $x$.

With a little bit of algebraic manipulation you get $$f = a\sqrt{1-\dfrac{b^2}{a^2}}$$

And this is how you get the coordinates of the focal points.

If you want to solve the problem without assuming that the foci are symmetrically placed on the major axis, you can assume arbitrary coordinates $(x_1, y_1)$ and $(x_2, y_2)$ for them, derive a general expression for $S$, and then solve for either $\frac{{\partial}S}{{\partial}h} = 0$ or $\frac{{\partial}S}{{\partial}k} = 0$, where $(h, k)$ are the coordinates of a point P on the ellipse.

  • $\begingroup$ A nice answer for the case if the ellipse isn't defined the way as in the WA :) +1. $\endgroup$
    – Rusurano
    Feb 23, 2023 at 8:47

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