JEEOlympiad

An ellipse is centred at the origin with its axes along the…

Question

An ellipse is centred at the origin with its axes along the coordinate axes. It passes through the points (2,3) and (4,1). Let l denote the length of its semi-latus rectum (l = b^2/a, where a is the length of the semi-major axis and b the semi-minor axis). If l^2 = m/n in lowest terms, find m+n.

✓ Verified answer: 79checked by our engine — not a guess

Step-by-step solution

Let the ellipse be x^2/A + y^2/B = 1, where A and B are the squared semi-axis lengths along x and y.
Through (2,3): 4/A + 9/B = 1.
Through (4,1): 16/A + 1/B = 1.
Let u = 1/A, v = 1/B: 4u + 9v = 1 and 16u + v = 1. From the second v = 1 - 16u; substitute: 4u + 9(1-16u) = 1 => 4u + 9 - 144u = 1 => -140u = -8 => u = 2/35, so A = 35/2. Then v = 1 - 32/35 = 3/35, so B = 35/3.
Since A = 35/2 > B = 35/3, the major axis is along x: a^2 = 35/2, b^2 = 35/3, a = sqrt(35/2).
Semi-latus rectum l = b^2/a = (35/3)/sqrt(35/2). Then
l^2 = (35/3)^2 / (35/2) = (1225/9)(2/35) = 2450/315 = 70/9.
m/n = 70/9, m+n = 79.

Final answer79

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