# Thread: Finding the equation of an ellipse

1. ## Finding the equation of an ellipse

Hello,
I have an assigned question which asks:

In the x-y plane let L be the line x = 1, let the point O be the origin,
and set L = 1

Find an equation for the set of points P = (x, y) such that the
ratio of the distances |OP|/|PM|= 1/2 where M is the point on
the line L that is closest to P. Note that from the class lecture on
the Kepler problem, you know that this curve is an ellipse. Since
one of the foci of the ellipse is at the origin, the equation may
not look like what you are used to.

Our lecture did not cover how to find the equation of the ellipse however if briefly described it and I have been looking in my text book for help but with no luck. If anyone could give me any advice and help me out I would really appreciate it.

2. ## Re: Finding the equation of an ellipse

Hello, lytwynk!

$\displaystyle \text{In the }xy\text{-plane, let }O\text{ be the origin and let }L\text{ be the line }x = 1.$

$\displaystyle \text{Find an equation for the set of points }P(x, y)$
$\displaystyle \text{such that the ratio of the distances: }\:\frac{|OP|}{|PM|} \,=\, \frac{1}{2}$
$\displaystyle \text{where }M\text{ is the point on line }L\text{ that is closest to }P.$

$\displaystyle \text{Note that from the class lecture on the Kepler problem, you know that this curve is an ellipse.}$
$\displaystyle \text{Since one of the foci of the ellipse is at the origin, the equation may not look like what you are used to.}$

The diagram looks like this:

Code:
        |     P     :M
|(x,y)* - - *(1.y)
|    /      :
|   /       :
|  /        :
| /         :
|/          :
- - * - - - - - + - - -
|O          1
|
$\displaystyle OP \:=\:\sqrt{x^2+y^2}$

$\displaystyle PM \:=\:1-x$

$\displaystyle \text{W\!e have: }\:\sqrt{x^2+y^2} \:=\:\tfrac{1}{2}(1-x) \quad\Rightarrow\quad 2\sqrt{x^2+y^2} \:=\:1 - x$

$\displaystyle \text{Square both sides: }\:4(x^2+y^2) \:=\:(1-x)^2 \quad\Rightarro\quad 4x^2 + 4y^2 \:=\:1-2x +x^2$

. . $\displaystyle 3x^2 + 2x + 4y^2 \:=\:1 \qiad\Rightarrow\quad 3\left(x^2 + \tfrac{2}{3}x\right) \,+\,4y^2 \:=\:1$

. . $\displaystyle 3\left(x^2 + \tfrac{2}{3}x + \tfrac{1}{9}\right) + 4y^2 \:=\:1 + \tfrac{1}{3} \quad\Rightarrow\quad 3\left(x + \tfrac{1}{3}\right)^2 + 4y^2 \:=\:\tfrac{4}{3}$

$\displaystyle \text{Divide by }\tfrac{4}{3}:\;\;\frac{3(x+\frac{1}{3})^2}{\frac{ 4}{3}} + \frac{4y^2}{\frac{4}{3}} \:=\:\frac{\frac{4}{3}}{\frac{4}{3}}$

$\displaystyle \text{And we have: }\:\frac{(x+\frac{1}{3})^2}{\frac{4}{9}} + \frac{y^3}{\frac{1}{3}} \:=\:1$

. . . $\displaystyle \begin{Bmatrix}\text{Center: }\:(\text{-}\frac{1}{3},\,0) \\ \\[-3mm] \text{Semimajor axis: }\: a \,=\,\frac{2}{3} \\ \\[-3mm] \text{Semiminor axis: }\:b \,=\,\frac{\sqrt{3}}{3}\end{Bmatrix}$

3. ## Re: Finding the equation of an ellipse

Thanks so much, this was a huge help.