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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.

Aug 10, 2020

#3
+115391
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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

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

$$x^2+0.2y^2=4 ​​ ​​$$

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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