Mathematics

Mathematics

Problem

A hyperbola, having the transverse axis of length 2sin⁡θ2 \sin \theta , is confocal with the ellipse 3x2+4y2=123x^2 + 4y^2 = 12 . Then, its equation is a. x2csc⁡2θ−y2sec⁡2θ=1x^2 \csc^2 \theta - y^2 \sec^2 \theta = 1 b. x2sec⁡2θ−y2csc⁡2θ=1x^2 \sec^2 \theta - y^2 \csc^2 \theta = 1 c. x2sin⁡2θ−y2cos⁡2θ=1x^2 \sin^2 \theta - y^2 \cos^2 \theta = 1 d. x2cos⁡2θ−y2sin⁡2θ=1x^2 \cos^2 \theta - y^2 \sin^2 \theta = 1
Intuitive explanation

The key is recognizing that "confocal" means the hyperbola and ellipse share the same two foci. First, find the foci of the ellipse using its standard form and eccentricity. Then use the constraint that the hyperbola's transverse axis has length 2sin⁡θ2\sin\theta to determine its semi-major axis aa. Since both curves have identical foci, their focal distances cc are equal—this relationship lets you find the hyperbola's semi-minor axis bb and write its equation.

Intuitive explanation

The key is recognizing that "confocal" means the hyperbola and ellipse share the same two foci. First, find the foci of the ellipse using its standard form and eccentricity. Then use the constraint that the hyperbola's transverse axis has length 2sin⁡θ2\sin\theta to determine its semi-major axis aa. Since both curves have identical foci, their focal distances cc are equal—this relationship lets you find the hyperbola's semi-minor axis bb and write its equation.

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