Q.If \(2y \cos \theta = x \sin \theta\) and \(2x \sec \theta - y \cosec \theta = 3\), then \(x^2 + 4y^2 = ?\) (a) 2 (b) 1 (c) 0 (d) None of the above
Answer: D
\(2y \cos \theta = x \sin \theta\) Or, \(2y \csc \theta = x \sec \theta\) Or, \(x \sec \theta - 2y \csc \theta = 0 \quad \text{——— (i)}\) Again, \(2x \sec \theta - y \csc \theta = 3 \quad \text{——— (ii)}\) Now, multiplying equation (i) by 2 and subtracting equation (ii), we get: \(-4y \csc \theta + y \csc \theta = -3\) Or, \(-3y \csc \theta = -3\) So, \(y = \frac{1}{\csc \theta} = \sin \theta\) Substituting \(y = \sin \theta\) into the equation \(2y \cos \theta = x \sin \theta\), we get: \(2 \sin \theta \cos \theta = x \sin \theta\) Or, \(x = 2 \cos \theta\) Therefore, \(x^2 + 4y^2 = (2 \cos \theta)^2 + 4 (\sin \theta)^2\) \(= 4 \cos^2 \theta + 4 \sin^2 \theta\) \(= 4 (\cos^2 \theta + \sin^2 \theta) = 4\)
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