Difficulty with an ellipse problem

[JosephRadkins]

Hello,

I recently was challenged to solve the problem which I am about to show you here, but I have reason to believe that it may be impossible. Is this question solvable, and if so, could anyone share some insight on how to begin? I have only a basic knowledge of Euclidean geometry and have never worked with ellipses before.

Many Thanks,
Joe

 

[soroban]

$\displaystyle \text{Hello, Joe!}$

$\displaystyle \text{If you're not familiar with ellipses, you are missing a }lot\text{ of information.}$


$\displaystyle \text{This ellipse has the equation: }\:\frac{x^2}{a^2} + \frac{y^2}{b^2} \:=\:1,\;\text{ with }c^2\:=\:a^2-b^2$

. . $\displaystyle \text{where: }\;a = CG,\;b = FC,\;c = BC.$


$\displaystyle \text{By definition, the sum of the distances from a point on the ellipse to the two foci}$
. . $\displaystyle \text{is a constant, namely, the major axis, }2a.$



$\displaystyle \text{In right triangle }FCG,\text{ we have: }\:FG^2 \:=\:CG^2 + FC^2 \quad\Rightarrow\quad FG^2 \:=\:a^2+b^2$

$\displaystyle \text{Since }b^2\:=\:a^2-c^2\text{, we have: }\;FG^2 \:=\:a^2 + (a^2-c^2)$

. . $\displaystyle \text{Hence: }\;FG^2 \:=\:2a^2 - c^2\;\;{\bf[1]}$


$\displaystyle \text{From the definition: }\;BF + DF \:=\:2a\quad\Rightarrow\quad 2\!\cdot\!\!BF \:=\:2a\quad\Rightarrow\quad BF \:=\:a$

. . $\displaystyle \text{and we know that: }\;c \:=\:BC$


$\displaystyle \text{Therefore, }{\bf[1]}\text{ becomes: }\;FG^2 \;=\;2\!\cdot\!\!BF^2 - BC^2$