<p>The equation of the pair of tangents drawn from the point (1, 2) to the ellipse \(3x^2 + 2y^2 = 5\) is</p>
Step-by-Step Solution
Key Concept: Use the standard formula SS₁ = T² for the pair of tangents from an external point to an ellipse, where S, S₁, and T are expressions involving the ellipse equation and the external point.
<p><strong>Solution:</strong> We use the formula for pair of tangents from an external point \(P(x_1, y_1)\) to the ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\):</p><p>\[SS_1 = T^2\]</p><p>where \(S = \frac{x^2}{a^2} + \frac{y^2}{b^2} - 1\), \(S_1 = \frac{x_1^2}{a^2} + \frac{y_1^2}{b^2} - 1\), and \(T = \frac{xx_1}{a^2} + \frac{yy_1}{b^2} - 1\)</p><p>For the ellipse \(3x^2 + 2y^2 = 5\), we rewrite as \(\frac{x^2}{5/3} + \frac{y^2}{5/2} = 1\), so \(a^2 = \frac{5}{3}\) and \(b^2 = \frac{5}{2}\)</p><p>From point (1, 2):</p><p>\(S = \frac{x^2}{5/3} + \frac{y^2}{5/2} - 1\)</p><p>\(S_1 = \frac{1}{5/3} + \frac{4}{5/2} - 1 = \frac{3}{5} + \frac{8}{5} - 1 = \frac{11}{5} - 1 = \frac{6}{5}\)</p><p>\(T = \frac{x}{5/3} + \frac{2y}{5/2} - 1 = \frac{3x}{5} + \frac{4y}{5} - 1\)</p><p>Applying \(SS_1 = T^2\) and simplifying yields: \(9x^2 - 4y^2 - 24xy + 40y + 30x - 55 = 0\)</p>
Correct Answer: A