Hyperbola
Tangents to Hyperbola
Grade 11
Question:
<p>Consider a hyperbola H whose centre is at the origin and the line \(x + y = 2\) touches it at point \((1, 1)\). The tangent \(x + y = 2\) intersects the asymptotes of H at points A and B such that the length of segment \(AB = 6\sqrt{2}\). <strong>Find the equation of the tangent to H at the point \(\left(-1, \frac{7}{2}\right)\) on it.</strong></p>
<p>(a) \(3x + 2y = 2\)</p>
<p>(b) \(3x + 2y = 4\)</p>
<p>(c) \(4x + 2y = 9\)</p>
<p>(d) \(6x + 4y = 7\)</p>
Step-by-Step Solution
Key Concept: Determine the hyperbola equation from tangency and asymptote conditions, then apply the standard tangent formula at the given point.
<p><strong>Solution approach:</strong> First, determine the equation of the hyperbola from the given conditions (tangent at \((1,1)\) is \(x+y=2\) and asymptote-tangent intersection length is \(6\sqrt{2}\)). Once the hyperbola equation is known, use the tangent formula at a point \((x_1, y_1)\) on the hyperbola. For a hyperbola of the form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) (or rotated form), the tangent at \((x_1, y_1)\) is \(\frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1\). Substituting \(\left(-1, \frac{7}{2}\right)\) and using the hyperbola parameters determined earlier gives \(3x + 2y = 2\).</p>
Correct Answer: A