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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.

 

Thanks in advance.

 Aug 10, 2020
 #3
avatar+111626 
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I am not very comfortable with conics but ...

If m is positive then this has to be an ellipse.

 

\(\frac{x^2}{1}+\frac{y^2}{1/m}=4\\ \frac{x^2}{4}+\frac{y^2}{4/m}=1\\\)

a=2,  b=sqrt(4/m)

 

a is 2 (units from the centre),     

which is less than the focal distance of 4 so this must be the minor axis.

So it goes through (2,0) and (-2,0)

 

If c is the focal distance and a and b are the lengths of the major and minor axis  (i.e.   2 and  sqrt(4/m))

then

\(c^2=|a^2-b^2|\\ 16=\frac{4}{m}-4\\ 20=\frac{4}{m}\\ m=\frac{1}{5}=0.2 \)

 

\( x^2+0.2y^2=4 ​​ ​​\)

 

--------------------------------

 

If m is negative then it will be an hyperbola.

\(x^2+my^2=4 \)

 

\(focal \;\;distance = \sqrt{1+m^2}\\ 4= \sqrt{1+m^2}\\ 1+m^2=16 m^2=15\\ m=-\sqrt{15}\\ m\approx -3.873\)

 

\(\bf{\text{So I get  }\;\;\;m=0.2 \;\;\;or\;\;\; m=-\sqrt{15}}\)

 

 

 

 

 

 

LaTex:

c^2=|a^2-b^2|\\
16=\frac{4}{m}-4\\
20=\frac{4}{m}\\
m=\frac{1}{5}

 

focal \;\;distance = \sqrt{1+m^2}\\
4= \sqrt{1+m^2}\\
1+m^2=16
m^2=15\\
m=-\sqrt{15}

 Aug 16, 2020

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