The 'foci' equation of an ellipse is
\(\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1 \)
a is the half the length of the horizontal axis and b is half the length of the vertical axis
(h,k) is the centre
If a is bigger than b then it will be more long, (the major axis will be horizontal)
If a is smaller than b then it will be more tall, (the major axis will be vertical)
So lets see what we have been given.
25x^2 +16 y^2 = 400
divide through by 400
\(\frac{x^2}{16}+\frac{y^2}{25}=1\\ \frac{x^2}{4^2}+\frac{y^2}{5^2}=1\\ \)
So the centre is (0,0)
It is a tall one.
The major axis will be from (-5,0) to (5,0)
The minor axis will be from (0,-4) to (0,4)
To find the focal points we use c where
\(c^2=|a^2-b^2|\\ c^2=|16-25|\\ c=3 \)
So the distance from the middle to the focal point is 3 units.
The foci will be at (0,-3) and (0,3)
We need to find the equation of the ellipse with
focal points (0,0) and (0,3) [I could have chosen (0,-3) if I had wanted to]
the centre will be (1,1.5)
c is the distance from the centre to the focal point, so c=1.5
It is just going to touch the other ellipse in one point and that point will be (0,5)
So the major axis b will be 5-1.5=3.5units
We have to find a
\(c^2=|a^2-b^2|\)
b >a so
\(c^2=b^2-a^2\\ a^2=b^2-c^2\\ a^2=3.5^2-1.5^2\\ a^2=10\\ a=\sqrt{10} \)
So the major axis is 2*3.5=7 units long
and
the minor axis will be 2sqrt10 which is approx 6.32 units long
For anyone interested: the equation of the second ellipse will be
\(\frac{x^2}{10}+\frac{(y-1.5)^2}{12.25}=1\)
