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So I was studying about ellipses in Polar Coordinates, and the book said

Let F be a fixed point, and l be a fixed line in a plane. Let e be a fixed positive number. The set of all points P in the plane such that $$\dfrac{|PF|}{|Pl|}=e$$ (a)an ellipse if, e $\lt 1$

In the proof, we let,

the focus at the origin, $|PF|=r, |Pl|=d-rcos\theta \quad$ (the equation of the directrix is x=d)

And by manupulating the equation $r=e(d-rcos\theta), $we have $$\dfrac {(x-h)^2}{a^2}+\dfrac{y^2}{b^2}=1\quad$$where $h=-\dfrac{e^2d}{1-e^2}\quad a^2=\dfrac{e^2d^2}{(1-e^2)^2}\quad b^2=\dfrac{e^2d^2}{1-e^2}$

And the book said the foci of an ellipse are at a distance c from the center where $$c^2=a^2-b^2=\dfrac{e^4d^2}{(1-e^2)^2}\quad c=-h$$

And my question is : If $h=c=0$, then we have an ellipse $\dfrac {x^2}{a^2}+\dfrac{y^2}{b^2}=1$ . and I thought if h=c=0, we were having a circle.

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Let me take an example. If we let$a=5,b=4 then c=3.$which means we have an ellipse that has foci (3,0),(-3,0). Can this ellipse also be shown by taking $h=-3 a=5,b=4$? – J.H. Sep 20 '13 at 11:05
Every circle is an ellipse. – mrf Sep 20 '13 at 11:06
@mrf then if $h=0,$ does it mean that $a=b=k(constant)?$ – J.H. Sep 20 '13 at 11:07
up vote 2 down vote accepted

The family of ellipses handled in the quoted passage was chosen specifically to have a simple equation in polar coordinates. Indeed, from the ratio $$ \frac{r}{d-r\cos\phi}=e $$ we easily get the polar equation $$ r=\frac{de}{1+e\cos\phi}\tag{1} $$ familiar to some of us from a course in celestial mechanics ;-)

Anyway, here the parameters that the user is free to choose are $d$ and $e\in[0,1)$. The other relevant coefficients: $a,b,h,c$ are then functions of $d$ and $e$, and cannot be chosen independently. For example, the center of an ellipse in this family is at the point $C=(h,0)=(-de^2/(1-e^2),0)$. As one of the focal points is fixed at the origin, $F_1=(0,0)$, the other focal point is then at $F_2=(2h,0)$, i.e. on the negative $x$-axis (assuming $d>0$).

So this family of ellipses does not include any of those with the familiar equation $$ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\tag{2} $$ because the foci of those ellipses are off the origin unless $a=b$. But if $a=b$, then we have a circle, i.e. an ellipse with eccentricity $e=0$. But if $e=0$, equation $(1)$ immediately implies $r=0$, i.e. a circle with radius zero. So the only ellipse of form $(2)$ that is also of form $(1)$ is the degenerate ellipse consisting of a single point.

In other words. Here an ellipse is formed by the loci of the points $P$ with the property that the ratio of their distances from a focal point and from the directrix is a constant. On a circle the distance to a focal point is also a constant, so the distance to the directrix must also be a constant. For this to happen either the circle collapses to a single point, or the directrix is at infinite distance. The latter case can be gotten here by a limiting process: letting $d\to\infty$ in such a way that $de$ remains a constant $R$. Then we must have $e\to0$, and $(1)$ becomes $r=R$ as expected.

Yet in other words. If one of the ellipses under discussion has its center at the origin, then $$ h=-\frac{de^2}{1-e^2}=0. $$ This implies that either $d=0$ or $e=0$. In either case equation $(1)$ gives $r=0$, i.e. a single point.

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Or still: $$b^2=-dh,\quad a^2=b^2/(1-e^2).$$ So if $h=0$, then also $a=b=0$. – Jyrki Lahtonen Sep 20 '13 at 13:41

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