Question #224025

If p is a parameter of positive numbers, show that all members of the family of ellipses x2/ (a2 + p) + y2 / (b2 + p) = 1 have the same foci.


1
Expert's answer
2021-08-18T12:39:20-0400

If the equation of an ellipse is x2m2+y2n2=1,\frac{x^2}{m^2}+\frac{y^2}{n^2}=1, the the foci are (c,0)(-c,0) and (c,0),(c,0), where c=m2n2.c=\sqrt{m^2-n^2}. In our case,


c=m2n2=(a2+p)2(b2+p)2=a2+p(b2+p)=a2b2.c=\sqrt{m^2-n^2}=\sqrt{(\sqrt{a^2+p})^2-(\sqrt{b^2+p})^2}\\ =\sqrt{a^2+p-(b^2+p)}=\sqrt{a^2-b^2}.


We conclude that the foci are (a2b2,0)(-\sqrt{a^2-b^2},0) and (a2b2,0).(\sqrt{a^2-b^2},0). Since the foci do not depend on the parameter p,p, the family of ellipses x2a2+p+y2b2+p=1\frac{x^2}{a^2 + p} + \frac{y^2}{b^2 + p} = 1 have the same foci.



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