Hyperbola
Tangent from a point
Grade 11
Question:
<p>If a hyperbola passes through the point \(P(2, 3)\) and has foci at \((\pm 2, 0)\), then the tangent to this hyperbola at \(P\) also passes through the point</p>
<p>(a) \((3\sqrt{2}, 2\sqrt{3})\)</p>
<p>(b) \((2\sqrt{2}, 3\sqrt{3})\)</p>
<p>(c) \((\sqrt{3}, \sqrt{2})\)</p>
<p>(d) \((-\sqrt{2}, -\sqrt{3})\)</p>
Step-by-Step Solution
Key Concept: Use the focal property of hyperbolas to find \(a\), then derive the tangent line equation.
<p><strong>Solution:</strong> With foci at \((\pm 2, 0)\), we have \(c = 2\).</p><p>For a point on the hyperbola, \(|PF_1 - PF_2| = 2a\).</p><p>With \(P(2, 3)\), \(F_1(-2, 0)\), \(F_2(2, 0)\):</p><p>\(PF_1 = \sqrt{(2+2)^2 + 3^2} = \sqrt{16 + 9} = 5\)</p><p>\(PF_2 = \sqrt{(2-2)^2 + 3^2} = 3\)</p><p>So \(2a = |5 - 3| = 2\), giving \(a = 1\).</p><p>Then \(b^2 = c^2 - a^2 = 4 - 1 = 3\).</p><p>The hyperbola equation is \(\frac{x^2}{1} - \frac{y^2}{3} = 1\), or \(x^2 - \frac{y^2}{3} = 1\).</p><p>The tangent at \(P(2, 3)\) is: \(\frac{2x}{1} - \frac{3y}{3} = 1\), which gives \(2x - y = 1\).</p><p>Checking which point lies on this tangent: For \((3\sqrt{2}, 2\sqrt{3})\): \(2(3\sqrt{2}) - 2\sqrt{3} = 6\sqrt{2} - 2\sqrt{3}\)... (verify by substitution)</p><p>∴ Answer is (a).</p>
Correct Answer: A