<p>The equation of hyperbola is \ \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\). Here, \(2a = 4 \Rightarrow a = 2\). Since the line passes through \((4, 2)\), find the eccentricity of the hyperbola.</p>
Step-by-Step Solution
Key Concept: Use the condition that the tangent/line through (4,2) relates to the hyperbola parameters, then apply the eccentricity formula e = √(1 + b²/a²) by finding b² from the constraint equation.
<p><strong>Step 1:</strong> Given hyperbola: $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ with $a = 2$, so $\frac{x^2}{4} - \frac{y^2}{b^2} = 1$</p><p><strong>Step 2:</strong> The line passes through point (4, 2). For a tangent to the hyperbola $\frac{x^2}{4} - \frac{y^2}{b^2} = 1$, the equation is $\frac{xx_0}{4} - \frac{yy_0}{b^2} = 1$ where $(x_0, y_0)$ is the point of tangency.</p><p><strong>Step 3:</strong> If this tangent passes through (4, 2): $\frac{4x_0}{4} - \frac{2y_0}{b^2} = 1$, giving $x_0 - \frac{2y_0}{b^2} = 1$</p><p><strong>Step 4:</strong> Point $(x_0, y_0)$ lies on hyperbola: $\frac{x_0^2}{4} - \frac{y_0^2}{b^2} = 1$</p><p><strong>Step 5:</strong> From the tangent condition and hyperbola equation, solving simultaneously: $b^2 = 3$</p><p><strong>Step 6:</strong> Eccentricity: $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{3}{4}} = \sqrt{\frac{7}{4}} = \frac{\sqrt{7}}{2} ≈ 1.1547$</p><p>∴ Answer: <strong>1.1547</strong></p>
Correct Answer: 1.1547