Analytic Geometry (high school): Why is the sum of the distances from any point of the ellipse to the two foci the major axis?

Question

I don't understand where that formula came from. Could someone explain? For example any point $(x,y)$ on the ellipse from the two foci $(-c,0)$ and $(c,0)$ is equal to $2a$ where $2a$ is the distance of the major axis. Where did this idea come from?

Answer 1

From the comments we know that the sum of the distances from a point $(x,y)$ on the ellipse to the two foci is a constant. Now, that's true for any point on the ellipse, so look at a point on the ellipse on the line connecting the two foci. In this case, we're looking at a point $(a,0)$ defining the semi-major axis, and the distance is $(a-c) + (a-(-c)) = 2a$.

Answer 2

Suppose that the foci of the ellipse are $(c,0)$ and $(-c,0)$, and that the major axis runs from $(-x,0)$ to $(x,0)$. Then the length of the major axis is $2x$. At the same time, the distance from $(x,0)$ to $(c,0)$ is $(x-c)$, and the distance from $(x,0)$ to $(-c,0)$ is $x-(-c) = x+c$. Then the sum of these distances is

$$ (x-c) + (x+c) = 2x$$

So the sum of the distances is precisely equal to the length of the major axis.

Answer 3

OK, imagine you have tacks at the points $(-c,0$ and $(c,0)$, which hold each end of a string of length $2 a$. We draw an ellipse by holding a pen taut against the string. The sum of the distances to a point on the ellipse from each of the tack points is

$$\sqrt{(x+c)^2+y^2} + \sqrt{(x-c)^2+y^2} = 2 a$$

The trick is to manage the algebra so that the derivation is readable. First, square both sides to get

$$(x-c)^2 + (x+c)^2 + 2 y^2 + 2 \sqrt{x^2+y^2+c^2+2 c x} \sqrt{x^2+y^2+c^2-2 c x} = 4 a^2$$

This simplifies a little to

$$x^2+y^2+c^2+\sqrt{(x^2+y^2+c^2)^2-4 c^2 x^2} = 2 a^2$$

Now we need to rid ourselves of this remaining square root by isolating it:

$$\begin{align}(x^2+y^2+c^2)^2-4 c^2 x^2 &= [2 a^2 - (x^2+y^2+c^2)]^2\\ &= 4 a^4 - 4 a^2 (x^2+y^2+c^2) + (x^2+y^2+c^2)^2 \end{align}$$

We have some fortuitous cancellation which leaves us with a quadratic. Rearrange to get

$$(a^2-c^2) x^2 + a^2 y^2 = a^2 (a^2-c^2)$$

or, in standard form:

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

Note that, for an ellipse, $a>c$. We interpret $a$ to be the semimajor axis, $c$ to be the focal length, and $b=\sqrt{a^2-c^2}$ is the semiminor axis.

Answer 4

As the center is the midpoint of the foci, so the center $O(0,0)$

As the foci lie on the major axis , so the equation of the major axis $y=0\implies$ the equation of the minor axis $x=0$

Now, if the length of the major, minor axes be $2a,2b$ respectively with eccentricity $=e$

So, the equation of the ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Any point $P$ on the ellipse can be $P(a\cos\theta,b\sin\theta)$

So, the distance between $(a\cos\theta,b\sin\theta), (ae,0)$ is

$\sqrt{(ae-a\cos\theta)^2+(b\sin\theta-0)^2}$ $=\sqrt{a^2e^2+a^2\cos^2\theta-2a^2e\cos\theta+a^2(1-e^2)(1-\cos^2\theta)}$ $=a(1-e\cos\theta)$ as $0\le e<1,-1\le \cos\theta\le 1$ and $b^2=a^2(1-e^2)$

Similarly, the distance between $(a\cos\theta,b\sin\theta), (-ae,0)$ is $=a(1+e\cos\theta)$