<p>A hyperbola has transverse axis of length \(2\sin\theta\) and is confocal with the ellipse \(3x^2 + 4y^2 = 12\). Then its equation is:</p>
<p>(a) \(x^2\csc^2\theta - y^2\sec^2\theta = 1\)</p>
<p>(b) \(x^2\sec^2\theta - y^2\csc^2\theta = 1\)</p>
<p>(c) \(x^2\sin^2\theta - y^2\cos^2\theta = 1\)</p>
<p>(d) \(x^2\cos^2\theta - y^2\sin^2\theta = 1\)</p>
Step-by-Step Solution
Key Concept: Confocal curves share the same foci. Use the relationship between semi-major axis, semi-minor axis, and distance to foci to find the equation.
<p><strong>Solution:</strong> The ellipse \(3x^2 + 4y^2 = 12\) can be written as \(\frac{x^2}{4} + \frac{y^2}{3} = 1\). Here \(a^2 = 4\) and \(b^2 = 3\), so \(c_e^2 = a^2 - b^2 = 1\), giving foci at \((\pm 1, 0)\). For a confocal hyperbola with transverse axis \(2a_h = 2\sin\theta\), we have \(a_h = \sin\theta\). Since the foci are at \((\pm c_h, 0)\) where \(c_h^2 = a_h^2 + b_h^2 = 1\), we get \(b_h^2 = 1 - \sin^2\theta = \cos^2\theta\). The hyperbola equation is \(\frac{x^2}{\sin^2\theta} - \frac{y^2}{\cos^2\theta} = 1\), or \(x^2\csc^2\theta - y^2\sec^2\theta = 1\).</p>
Correct Answer: a