UPSC NDA I 2022 Mathematics Solved Paper

Concept:The distance between the foci of an ellipse is given by $2ae$, where $a$ is the length of the semi-major axis and $e$ is the eccentricity.For an ellipse of the form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with $a > b$, the eccentricity is $e = \sqrt{1 - \frac{b^2}{a^2}}$.Explanation:Step 1: Rewrite the given equation $x^2 + 2y^2 = 1$ in standard form.Divide both sides by 1: $\frac{x^2}{1} + \frac{y^2}{\frac{1}{2}} = 1$.Hence, $a^2 = 1$ and $b^2 = \frac{1}{2}$. Since $a^2 > b^2$, the major axis is along the $x$-axis.Step 2: Calculate the eccentricity $e$.$e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{1/2}{1}} = \sqrt{1 - \frac{1}{2}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.Step 3: Find the distance between the foci.The distance between the two foci of an ellipse is $2ae$.Here, $a = \sqrt{1} = 1$, so $2ae = 2 \times 1 \times \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$.Answer:The distance between the foci is $\sqrt{2}$, which corresponds to option B.