UPSC NDA I and II 2020 Math Solved Paper

Concept:For any point on an ellipse, the sum of its distances to the two foci is constant and equal to the length of the major axis.Explanation:Step 1: Rewrite the ellipse equation in standard form.$25x^2 + 16y^2 = 400 \implies \frac{x^2}{16} + \frac{y^2}{25} = 1$Here $a^2 = 16$, $b^2 = 25$, so $a = 4$, $b = 5$ and $b > a$. Hence the major axis is along the $y$-axis.Step 2: Find the eccentricity $e$.$e = \sqrt{1 - \frac{a^2}{b^2}} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}$Step 3: Find the foci coordinates.For $b > a$, the foci are $(0, \pm be) = \left(0, \pm 5 \cdot \frac{3}{5}\right) = (0, \pm 3)$.Thus the given points $Q(0,3)$ and $R(0,-3)$ are exactly the two foci of the ellipse.Step 4: Apply the property of ellipse.For any point $P$ on the ellipse, $PQ + PR =$ sum of focal distances = length of major axis = $2b = 2 \times 5 = 10$.(Alternatively, choose a convenient point $P$, e.g., $(4,0)$, and compute distances: $PQ = \sqrt{(4-0)^2+(0-3)^2} = 5$, $PR = \sqrt{(4-0)^2+(0+3)^2} = 5$, sum = $10$.)Answer:$10$