Hyperbola
Tangent to Hyperbola
Grade 11
Question:
<p>Let the equation of hyperbola be \(\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1\). It passes through \((4, 6)\) and the eccentricity is \(2\). Find the equation of the tangent to the hyperbola at \((4, 6)\).</p>
<p>(1) \(2x - y + 2 = 0\)</p>
<p>(2) \(2x + y - 2 = 0\)</p>
<p>(3) \(2x - y - 2 = 0\)</p>
<p>(4) \(x - 2y + 2 = 0\)</p>
Step-by-Step Solution
Key Concept: Use the eccentricity relation e² = 1 + b²/a² to find the relationship between a and b, then substitute the point (4,6) to determine the hyperbola equation, and finally apply the tangent formula at a point on the hyperbola.
<p><strong>Step 1:</strong> Use eccentricity relation for hyperbola.</p><p>Given: e = 2, so e² = 4</p><p>For hyperbola: e² = 1 + b²/a²</p><p>∴ 4 = 1 + b²/a² → b²/a² = 3 → b² = 3a²</p><p><strong>Step 2:</strong> Use the point (4, 6) to find a².</p><p>Hyperbola passes through (4, 6):</p><p>16/a² - 36/(3a²) = 1</p><p>16/a² - 12/a² = 1</p><p>4/a² = 1 → a² = 4</p><p>∴ b² = 3(4) = 12</p><p><strong>Step 3:</strong> Write the hyperbola equation.</p><p>x²/4 - y²/12 = 1</p><p><strong>Step 4:</strong> Find tangent at (4, 6).</p><p>Tangent equation at point (x₀, y₀): xx₀/a² - yy₀/b² = 1</p><p>At (4, 6): 4x/4 - 6y/12 = 1</p><p>x - y/2 = 1</p><p>∴ <strong>2x - y - 2 = 0</strong></p>
Correct Answer: C