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Conic Sections
Ellipse
Hard

Question

The length of the chord of the ellipse x225+y216=1\frac{x^2}{25}+\frac{y^2}{16}=1, whose mid point is (1,25)\left(1, \frac{2}{5}\right), is equal to :

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Solution

1. Introduction and Key Concept: Equation of a Chord with a Given Midpoint

The problem asks us to determine the length of a chord of the ellipse x225+y216=1\frac{x^2}{25}+\frac{y^2}{16}=1, given that its midpoint is (1,25)\left(1, \frac{2}{5}\right). A fundamental and highly efficient approach for problems involving chords and their midpoints in conic sections is to utilize the T=S1T=S_1 formula.

For any general conic section represented by the equation S=Ax2+Bxy+Cy2+Dx+Ey+F=0S = Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0, the equation of a chord whose midpoint is (x1,y1)(x_1, y_1) is given by the formula: T=S1T = S_1 Let's define TT and S1S_1 specifically for an ellipse in the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2

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