<p>If the line \(y = mx + 7\sqrt{3}\) is normal to the hyperbola \(\dfrac{x^2}{24} - \dfrac{y^2}{18} = 1\), then a value of \(m\) is __________ (up to four decimal places).</p>
Step-by-Step Solution
Key Concept: For a line y = mx + c to be normal to a hyperbola, it must satisfy the condition that the slope m and intercept c are related through the normal line equation. Use the property that for hyperbola x²/a² - y²/b² = 1, a normal at point (x₀, y₀) has slope -b²x₀/(a²y₀), and substitute the given constraint.
<p><strong>Step 1:</strong> For hyperbola x²/24 - y²/18 = 1, we have a² = 24, b² = 18.</p><p><strong>Step 2:</strong> For a normal line y = mx + c to the hyperbola x²/a² - y²/b² = 1, the condition is:<br/>c = (a² + b²)m³/b²</p><p><strong>Step 3:</strong> Here c = 7√3, a² + b² = 24 + 18 = 42, b² = 18. Substitute:<br/>7√3 = 42m³/18<br/>7√3 = 7m³/3<br/>m³ = 3√3</p><p><strong>Step 4:</strong> m = ∛(3√3) = ∛(3^(3/2)) = 3^(1/2) = √3<br/>m = 1.7321 (approximately)</p><p><strong>Verification:</strong> 42(√3)³/18 = 42(3√3)/18 = 126√3/18 = 7√3 ✓</p><p>∴ Answer: <strong>1.7321</strong> or <strong>√3</strong></p>
Correct Answer: 1