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What is the distance between the foci of the ellipse x2 + 2y2 = 1 ?

A.
1
B.
\(\sqrt 2\)
C.
2
D.
\(2\sqrt 2\)

Solution:

Concept:The general equation of an ellipse is written as: \(\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) where a and b are the lengths of the semi-major and semi-minor axes respectively.Distance between foci is = 2ae, where e = \(\displaystyle \sqrt{1-\frac{b^2}{a^2}} \) Calculation:Given:The equation of the ellipse is x2 + 2y2 = 1? \(\displaystyle \frac{x^2}{1}+\frac{y^2}{\frac{1}{2}}=1\)? a2 = 1, b2 = \(\displaystyle \frac{1}{2}\)e = \(\displaystyle \sqrt{1-\frac{b^2}{a^2}} \) = \(\displaystyle \sqrt{1-\frac{1}{2}} \) = \(\displaystyle \frac{1}{\sqrt2}\)Distance between foci is = 2ae = 2 × 1 × \(\displaystyle \frac{1}{\sqrt2}\) = \(\sqrt 2\)? Distance between foci is \(\sqrt 2\)

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