ellipse triangle issue

[TurboStevie]

Hello all,
This problem is taken from a test which my boyfriend has to take for his job. I am not very geometry savvy. I excelled more at statistics. Despite my geometric shortcomings, I'm fairly certain that there is not enough information in the diagram to prove any of the optional answers. My best go at it would be CD thinking that CFD could be an equilateral triangle, but I can't find enough info in the diagram to prove it. Feel free to ignore erased markings and blue highlighter. I was scribbling all over the darn thing.
Thank you for any help!

 

[Steven G]

Hello all,
This problem is taken from a test which my boyfriend has to take for his job. I am not very geometry savvy. I excelled more at statistics. Despite my geometric shortcomings, I'm fairly certain that there is not enough information in the diagram to prove any of the optional answers. My best go at it would be CD thinking that CFD could be an equilateral triangle, but I can't find enough info in the diagram to prove it. Feel free to ignore erased markings and blue highlighter. I was scribbling all over the darn thing.
Thank you for any help!View attachment 4861

CF =a. And AE =a. So....
I used the fact that the general equation of the ellipse you gave us is x^2/a^2 + y^2/b^2 =1. You need to understand what a means in the equation.

 

[HallsofIvy]

The ellipse \(\displaystyle \frac{x^2}{a^2}+ \frac{y^2}{b^2}= 1\) has "vertices" at (a, 0), (-a, 0), (0, b), and (0, -b). Also any ellipse has the property that the total length from one focus, to any point on the ellipse, back to the other focus, is constant. Imagine taking that length from focus F to (a, 0) then back to (-a, 0). If the two foci are at (e, 0) and (-e, 0), then the distance from (e, 0) to (a, 0) is a- e. The distance from (a, 0) back to (-e, 0) is a- (-e)= a+ e. The total distance is (a- e)+ (a+ e)= 2a. That means that the distance from focus (e, 0) to (0, b) back to (-e, 0) is also 2e. If I am correct that "F" is one of the foci and "C" is (0, b) (it's hard to read your picture!) then CF has length a which is the length of AE.

 

[TurboStevie]

Thank you to the both of you!

The ellipse \(\displaystyle \frac{x^2}{a^2}+ \frac{y^2}{b^2}= 1\) has "vertices" at (a, 0), (-a, 0), (0, b), and (0, -b). Also any ellipse has the property that the total length from one focus, to any point on the ellipse, back to the other focus, is constant. Imagine taking that length from focus F to (a, 0) then back to (-a, 0). If the two foci are at (e, 0) and (-e, 0), then the distance from (e, 0) to (a, 0) is a- e. The distance from (a, 0) back to (-e, 0) is a- (-e)= a+ e. The total distance is (a- e)+ (a+ e)= 2a. That means that the distance from focus (e, 0) to (0, b) back to (-e, 0) is also 2e. If I am correct that "F" is one of the foci and "C" is (0, b) (it's hard to read your picture!) then CF has length a which is the length of AE.

HallsofIvy! This was great! Thank you! This was broken down wonderfully. Even though I'm sure I'm at some freakishly remedial level for a full blown college graduate adult, I've always loved math. Thanks again!