+0

# Ahhh 3rd repost, seems past answers I couldn’t understand and were incorrect (not angry)

0
45
2

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.

Mar 7, 2021

### 2+0 Answers

#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

Mar 7, 2021
#2
+1

x2  + my2 = 4

x2 /4  +  m y2 /4  = 1

x/ 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

Mar 7, 2021