Suppose A, B, C and D are points on an ellipse such that segments AB and CD intersect at a focus F. Given that AF = 3, CF = 4 and BF = 5, what is DF?
This is in our subject precalculus. Our topic is about ellipse, we are restricted to use the three trigonometric ratios(sine, cosine, tangent).
I tried to solve the problem by assuming that the segment AB is the major axis and the segment CD is the latus rectum since the endpoints of major axis and latus rectum are points on an ellipse and they both intersect at the focus.
As you can see in the figure, $AF=3$ $BF=5$ $CF=4$, since segment $CD$ is a latus rectum, $CF$ should be equal to $DF$. meaning $DF$ should also be equal to $4$.
But when I checked if it is right, given that latus rectum is equal to $\frac{2b^2}{a}$. First to solve for $b$, $b^2=a^2-c^2$, we can get the value of $c$ and $a$ in the figure, $a=4$ and $c=1$
$b^2=4^2-1^2$
$b=\sqrt{15}$
now to confirm if $DF=4$, I solve for the half of latus rectum $\frac{1}{2}\cdot\frac{2b^2}{a}$
$\frac{1}{2}\cdot\frac{2\sqrt{15}^2}{4}=3.75$
$DF$ is not equals to $4$ :<
Now I'm having a hard time thinking another way to solve the problem considering the restrictions that are given(not allowed to use trigonometric ratios as mentioned above). Can you give me some clue or tips on how to solve this problem? Thank you!