Answer:
AKPOS/ACGF = 28√3 -48 ≈ 0.497423
Step-by-step explanation:
Let's assume the square is a unit square. Then the height of the triangle is ...
1 +(√3)/2 = h
and half the base of the triangle is ...
h/√3 = (√3)/3 +1/2 = b/2
The area of the triangle is the product of these:
A = (1/2)bh
= (2 +√3)/2 × (3 +2√3)/6
= (6 +3√3 +4√3 +6)/12 = (12 +7√3)/12
So, the ratio of the area of the square to that of the triangle is the inverse of this, or ...
(square area)/(triangle area) = 12/(12+7√3)
(square area)/(triangle area) = 28√3 -48
I don't know if we can find the foci of this ellipse, but we can find the centre and the vertices. First of all, let us state the standard equation of an ellipse.
(If there is a way to solve for the foci of this ellipse, please let me know! I am learning this stuff currently.)

Where

is the centre of the ellipse. Just by looking at your equation right away, we can tell that the centre of the ellipse is:

Now to find the vertices, we must first remember that the vertices of an ellipse are on the major axis.
The major axis in this case is that of the y-axis. In other words,
So we know that b=5 from your equation given. The vertices are 5 away from the centre, so we find that the vertices of your ellipse are:

&

I really hope this helped you! (Partially because I spent a lot of time on this lol)
Sincerely,
~Cam943, Junior Moderator
It is A because you are adding 3 to in if you were to subtract then it will go down not up.