Let $m$ be a constant not equal to $0$ or $1.$ Then the graph of \[x^2 + my^2 = 4\]is a conic section with two foci. Find all values of $m$ such that the foci both lie on the circle $x^2+y^2=16.$ Enter all possible values of $m,$ separated by commas.
Oh I just found out my mistake, so when I tried to find the equation of Foci, which was a^2 -b^2=c^2
all I did was a^2-b^2=c!
btw I got -1/3, 1/5
+37158 x2 + my2 = 4
x2 /4 + m y2 /4 = 1
x2 / 4 + y2 / ( m-1 *4) = 1
c^2 = a^2 - b^2 we want the FOCI to be +-4 to be on the circle of radius 4 ( c is the focus)
42 = m-1 *4 - 4
20 = m-1 *4
5 = m^-1
m = .2
x^2 + .2 y^2 = 4 ( I think !) I also got - 1/3 ....but I do not think that is an ellipse conic section
See graph:
https://www.desmos.com/calculator/n8wecwwstr
