I'm having hard time in learning the Ellipse. I'm using The Elements of Coordinate Geometry by S.L. Loney and his derivation of equation to an ellipse goes like this
Let $ZK$ be the directrix, $S$ the focus, and let $SZ$ be perpendicular to the directrix
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There will be a point $A$ on $SZ$ such that $$ SA = e\cdot AZ$$ Since, $e\lt1, $ there will be another point $A'$, in $ZS$ produced, such that $$ SA' = e\cdot A'Z$$ Let the the length of $AA'$ be called $2a$, and let $C$ be the middle point of $AA'$. Adding equations so formed, we have $$ 2a = AA' = e(AZ+A'Z)=2\cdot e \cdot CZ$$ i.e. $$ CZ= \frac{a}{e} $$ Subtracting the equations, we have $$ e(A'Z-AZ)=SA'-SA=(SC+CA')-(CA-CS)$$ i.e. $$e\cdot AA'=2CS$$ $$ CS = a\cdot e.$$ Let $C$ be the origin, $CA'$ the axis of x and a line through C perpendicular to AA' the axis of Y.
Let P be any point on the curve, whose coordinates are $x$ and $y$, and let PM be the perpendicular upon the directrix, and PN the perpendicular upon AA'.
The focus S is the point $(-ae,)$. The relation $SP^2= e^2 \cdot PM^2= e^2 \cdot ZN^2$ then gives $$ (x+ae)^2 +y^2 = e^2(x+\frac{a}{e})^2$$ $$x^2(1-e^2) +y^2 = a^2 (1-e^2)$$ $$ \frac{x^2}{a^2} + \frac{y^2}{a^2(1-e^2)} =1$$ if we put $$b= a\sqrt{1-e^2}$$ then we have $$ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1.$$
. As you can see, when he says Let C be the origin and this simplifies everything and because this only that we get a nice form like this $$ \frac{x^2}{a^2} + \frac{y^2}{b^2} =1$$ I want to know how to find the equation to an ellipse if the center $C$ were to be at $(h,k)$.
Here is my try : Consider the coordinates of $C$ to be $(h,k)$ and the coordinates of $S$ to be $(p,q)$, since $CS$ is $a~e$, therefore $$ (p-h)^2 + (q-k)^2 = a^2~e^2$$ and similarly coordinates of $Z$ can be assumed to be $(c,d)$ and again $$ (c-h)^2 + (d-k)^2 =1 .$$ Now, let's turn to our sample point $P(x,y)$, by the definition of the ellipse: $$ SP^2 = e^2 PM^2 $$ $$ (p-x)^2 + (q-y)^2 = e^2 NZ^2 $$ and now comes the problem, I can't write $N$ as $(x,0)$ because $ZCZ'$ is no longer the $x-$axis. Why only in ellipse that we are getting this problem? It was so easy with circles and parabola but as soon as I have been introduced to ellipse I started getting problems. You, the equation of parabola $y^2 = 4ax$ tell us the opening is on the positive $x-$axis and it lies on $x-$axis and $x^2 = 4ay$ is just telling us that now everything is on $y-$ axis but in the case of ellipses it's we who have to determine how the eclipse would look like by comparing $a$ and $b$ (semi major and minor axis lengths). If we are given an equation like $$ \frac{x^2}{a^2} + \frac{y^2}{b^2} =1 $$ and said that find the latus rectum then which formula should we use $\frac{2b^2}{a}$ or $\frac{2a^2}{b}$ (if in question it is given only $a$ and $b$ and not told which one is greater).
Thank you. Any help will be much appreciated.
If $f(x,y)=0$ is the equation of any set in $\mathbb{R}^2$, then $f(x-h,y-k)=0$ is the equation of the translation of that set by the vector $(h,k)$. Therefore an ellipse with center $(h,k)$ can be written as $$\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1.$$