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A hyperbola has a transverse axis of length 2 sin θ and is confocal with the ellipse 3x² + 4y² = 12. Its equation is
- x² cosec² θ - y² sec² θ = 1
- x² sec² θ - y² cosec² θ = 1
- x² sin² θ - y² cos² θ = 1
- x² cos² θ - y² sin² θ = 1
Correct answer: x² cosec² θ - y² sec² θ = 1
Solution
3x^2+4y^2=12 gives c^2=4-3=1, foci (+-1,0). Confocal hyperbola with transverse semi-axis sin(theta) has A=sin^2(theta), and c^2=A+B=1 gives B=cos^2(theta): x^2 cosec^2(theta) - y^2 sec^2(theta) = 1.
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