How do different definitions of ellipse translate to the same thing?

Question

There are 2 definitions of an ellipse that I know.

One definition goes:

The locus of a point moving in a plane such that the ratio of its distances from a fixed line (directrix) and a fixed point (focus) is a constant and less than 1.

Another goes:

The locus of a point moving in a plane such that the sum of its distances from two fixed point is a constant.

Now my question is how is it that these two definitions, which seem quite a lot different from each other, define the same thing - ellipse? One more thing I'd like to know is that in the first definition, only one fixed point and only one fixed line are mentioned. Whereas in the second one, there's no mention of the fixed line and two fixed points are mentioned. And one of the properties of ellipse is that there are two focii and two directrices. Now all this is quite confusing to me. I know that they all define the same thing, ellipse, but I don't know how exactly ( heck, not even roughly! :D ) are these definitions equivalent.

Answer 1

For simplicity, pick a point $(f,0)$ on the $x$-axis and a vertical line $x=L$ with $L>f>0$.

Let us consider all points $(x,y)$ such that this ratio is a constant $e<1$:

$$\frac{{\rm dist}({\rm point}\,(x,y),\,{\rm point}\,(f,0))}{{\rm dist}({\rm point}\,(x,y),\,{\rm line}\,x=L)}=\frac{\sqrt{(x-f)^2+y^2}}{|x-L|}=e.$$

Squaring yields

$$x^2-2fx+f^2+y^2=e^2x^2-2e^2Lx+e^2L^2$$

$$\iff (1-e^2)x^2+2(e^2L-f)x+y^2=e^2L^2-f^2.$$

We can slide the point $(f,0)$ and line $x=L$ along the $x$-axis without changing the shape or size of the resulting curve; these only affect its location. Thus we can without loss of generality move them so that $f=e^2L$. (Suppose $f$ and $L$ are distance $d$ apart: then simply solve $f=e^2(f+d)$ for $f$.)

At this point we have $(1-e^2)x^2+y^2=e^2L^2-f^2$. Dividing, we obtain

$$\frac{1-e^2}{e^2L^2-f^2}x^2+\frac{1}{e^2L^2-f^2}y^2=1,$$

which is of the form $\square x^2+\square y^2=1$, an ellipse. Conversely, let

$$b^2=e^2L^2-f^2,\qquad a^2=\frac{b^2}{1-e^2}$$

(note the condition $f=e^2L$ forces $f,e,L$ to have exactly two degrees of freedom), hence

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

is the general form of ellipse obtained. Working backwards (solving for $e,f,L$ in terms of $a,b$ under the condition $f=e^2L$) yields the focus and directrix of an ellipse. Note these ellipses are specifically those with center $(0,0)$ that are longer horizontally than vertically; every other ellipse can be obtained by performing rigid motions on the situation we've analyzed.

Similar work shows that $e=1$ defines a parabola, and $e>1$ a hyperbola.

If we view the equation $(x/a)^2+(y/b)^2=1$ as a midpoint between the two definitions, this answers half your question: how the first definition always generates an ellipse, and how every ellipse falls under the first definition. lab's answer connects this midpoint to the second definition.

With your powers combined, I am Together, these show that the two definitions are equivalent.

Answer 2

WLOG, we can assume the equation of the ellipse to be $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$ and any point on it can be written as $P(a\cos t,b\sin t)$

Let the foci be $S(ae,0), S'(-ae,0)$

So, $|PS|=\sqrt{(ae-a\cos t)^2+(0-b\sin t)^2}$ $=\sqrt{(ae-a\cos t)^2+a^2(1-e^2)(1-\cos^2t)}$

$=a(1-e\cos t)$ as $e\cos t<1$ $0<e<1$ and $-1\le \cos t\le1$

Similarly, $PS'=a(1+e\cos t)$

So, the sum of the distances is $2a$ which is independent of $t,$ hence constant and in fact equals to the length of the major axis.

Conversely,

WLOG we can assume the fixed points be $S(a,0)$ and $S'(-a,0)$ where $a>0$

So, $\sqrt{(x-a)^2+(y-0)^2}+\sqrt{(x+a)^2+(y-0)^2}=2c$(say)

Observe that the shortest distance between $S(a,0)$ and $S'(-a,0)$ is $a-(-a)=2a$

So, $2c$ must be $>2a\implies c>a$

On sqauring & re-arrangment we get , $$\frac {x^2}{c^2}+\frac{y^2}{c^2-a^2}=1$$ which is a standard equation of an ellipse as $c>a$