Hyperbola
Tangent to Hyperbola
Grade 11
Question:
<p>If the eccentricity of the standard hyperbola passing through the point \((4, 6)\) is 2, then the equation of the tangent to the hyperbola at \((4, 6)\) is:</p>
<p>\(x - 2y + 8 = 0\)</p>
<p>\(2x - 3y + 10 = 0\)</p>
<p>\(2x - y - 2 = 0\)</p>
<p>\(3x - 2y = 0\)</p>
Step-by-Step Solution
Key Concept: For a hyperbola with eccentricity e=2, use e² = 1 + b²/a² to find the relationship between a and b, then use the point (4,6) to determine the specific hyperbola equation before finding the tangent.
<p><strong>Step 1:</strong> Use eccentricity condition. For hyperbola x²/a² - y²/b² = 1, we have e² = 1 + b²/a²</p><p>Given e = 2: 4 = 1 + b²/a² → b²/a² = 3 → b² = 3a²</p><p><strong>Step 2:</strong> Hyperbola equation becomes x²/a² - y²/(3a²) = 1</p><p><strong>Step 3:</strong> Point (4,6) lies on hyperbola:</p><p>16/a² - 36/(3a²) = 1</p><p>16/a² - 12/a² = 1</p><p>4/a² = 1 → a² = 4, b² = 12</p><p><strong>Step 4:</strong> Hyperbola equation: x²/4 - y²/12 = 1</p><p><strong>Step 5:</strong> Tangent at (4,6) using formula (xx₁)/a² - (yy₁)/b² = 1:</p><p>(4x)/4 - (6y)/12 = 1</p><p>x - y/2 = 1</p><p>2x - y = 2 or <strong>2x - y - 2 = 0</strong></p><p>∴ Answer: C</p>
Correct Answer: C