+0  
 
0
45
2
avatar

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
 #1
avatar
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
avatar+31208 
+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

36 Online Users

avatar
avatar
avatar
avatar