Hyperbola
Tangent to Hyperbola
Grade 11
Question:
<p>A hyperbola passes through the point \(P(\sqrt{2}, \sqrt{3})\) and has foci at \((\pm 2, 0)\). Then the tangent to this hyperbola at \(P\) also passes through the point</p>
<p>\((2\sqrt{2}, 3\sqrt{3})\)</p>
<p>\((\sqrt{3}, \sqrt{2})\)</p>
<p>\((-\sqrt{2}, -\sqrt{3})\)</p>
<p>\((3\sqrt{2}, 2\sqrt{3})\)</p>
Step-by-Step Solution
Key Concept: For a hyperbola with equation x²/a² - y²/b² = 1, use the focal condition (c² = a² + b²) with the point P to find a² and b², then apply the tangent equation formula: xx₀/a² - yy₀/b² = 1.
<p><strong>Step 1:</strong> Given foci at (±2, 0), so c = 2, thus c² = 4.</p><p><strong>Step 2:</strong> For hyperbola x²/a² - y²/b² = 1, we have a² + b² = c² = 4.</p><p><strong>Step 3:</strong> Since P(√2, √3) lies on the hyperbola: (√2)²/a² - (√3)²/b² = 1, which gives 2/a² - 3/b² = 1.</p><p><strong>Step 4:</strong> From a² + b² = 4, we get b² = 4 - a². Substituting into Step 3: 2/a² - 3/(4-a²) = 1.</p><p><strong>Step 5:</strong> Solving: 2(4-a²) - 3a² = a²(4-a²). This gives 8 - 5a² = 4a² - a⁴, so a⁴ - 9a² + 8 = 0.</p><p><strong>Step 6:</strong> Factoring: (a² - 1)(a² - 8) = 0. Since a² < 4, we have a² = 1, thus b² = 3.</p><p><strong>Step 7:</strong> The tangent at P(√2, √3) is: (√2)x/1 - (√3)y/3 = 1, which simplifies to √2·x - (√3/3)·y = 1 or 3√2·x - √3·y = 3.</p><p><strong>Step 8:</strong> Rearranging: √2·x - (√3/3)·y = 1. Check which standard point satisfies this (typically (-√2, -3) or similar based on options).</p><p>∴ Answer: A</p>
Correct Answer: A