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.

Guest Mar 7, 2021

#1**0 **

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

Guest Mar 7, 2021

#2**+1 **

x^{2} + my^{2} = 4

x^{2 }/4 + m y^{2} /4 = 1

x^{2 }/ 4 + y^{2 }/ ( 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)

4^{2 }= 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

ElectricPavlov Mar 7, 2021