Question

What are the foci of the ellipse given by the equation `225x^2+144y^2=32,400`?

Answer 1

Foci of the ellipse are `(0,-9)` and `(0,9)`

Explanation:

`225x^2+144y^2=32400` can be written as

`x^2/(32400/225)+y^2/(32400/144)=1`

or `x^2/144+y^2/225=1`

or `x^2/12^2+y^2/15^2=1`

Hence major axis is `15xx2=30` - along `y`-axis and minor axis is `12xx2=24` along `x`-axis and hence foci too will be along `y`-axis.

and eccentricity is given by `12^2=15^2(1-e^2)` i.e.

eccentricity is `e=sqrt(1-12^2/15^2)=3/5`

and hence foci are `(0,-15xx3/5)` and `(0,15xx3/5)`

i.e. `(0,-9)` and `(0,9)`

graph{(225x^2+144y^2-32400)(x^2+(y+9)^2-0.16)(x^2+(y-9)^2-0.16)=0 [-40, 40, -20, 20]}