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Geometry Level 3

If the foci of the ellipse x 2 25 + y 2 9 = 1 \large \frac{x^2}{25}+\frac{y^2}{9}=1 are ( 4 , 0 ) (4,0) and ( − 4 , 0 ) (-4,0) , then for real number λ \lambda , the foci of the ellipse x 2 25 + λ 2 + y 2 9 + λ 2 = 1 \large \frac{x^2}{25+ \lambda^2}+\frac{y^2}{9+\lambda^2}=1 are:

( λ 2 , 0 ) , ( − λ 2 , 0 ) (\lambda^2,0),(-\lambda^2,0) ( 0 , 4 ) , ( 0 , − 4 ) (0,4),(0,-4) ( 0 , 0 ) , ( 0 , 0 ) (0,0),(0,0) ( 4 + λ 2 , 0 ) , ( − 4 + λ 2 , 0 ) (4+\lambda^2,0),(-4+\lambda^2,0) ( 4 , 0 ) , ( − 4 , 0 ) (4,0),(-4,0) ( λ , 0 ) , ( − λ , 0 ) (\lambda,0),(-\lambda,0)

Vishnu Bhagyanath by Vishnu Bhagyanath , 23, India

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1 solution

Sean Sullivan
Jul 28, 2015

The foci of an ellipse x 2 a 2 + y 2 b 2 = 1 \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 where a > b a>b are ( ± a 2 − b 2 , 0 ) \left(\pm\sqrt{a^{2}-b^{2}},0\right) , so for the second equation we have foci at ( ± 25 + λ 2 − ( 9 + λ 2 ) , 0 ) = ( ± 16 , 0 ) \left(\pm\sqrt{25+\lambda^{2}-(9+\lambda^{2})},0\right)=\left(\pm\sqrt{16},0\right) hence the answer is ( 4 , 0 ) , ( − 4 , 0 ) \boxed{(4,0),(-4,0)}

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