Hyperbola
Properties and Tangency
Grade 11
Question:
<p>Consider an ellipse \(\frac{x^2}{36} + \frac{y^2}{18} = 1\). There is a hyperbola whose one asymptote is the major axis of the given ellipse. If the eccentricity of the given ellipse and hyperbola are reciprocal to each other, both have the same centre, and both touch each other in the first and third quadrants, find the focus of the hyperbola.</p>
<p>(a) \(\left(\frac{3\sqrt{3}}{2}, \frac{3\sqrt{3}}{2}\right)\)</p>
<p>(b) \(\left(\frac{3}{2}, \frac{3}{2}\right)\)</p>
<p>(c) \((3\sqrt{2}, 3\sqrt{2})\)</p>
<p>(d) \([3(2^{3/4}), 3(2^{3/4})]\)</p>
Step-by-Step Solution
Key Concept: Use the relationship between eccentricities and the tangency condition to determine the hyperbola's parameters, then find its focus.
<p><strong>Solution:</strong> For the ellipse \(\frac{x^2}{36} + \frac{y^2}{18} = 1\): \(a^2 = 36, b^2 = 18\), so \(c^2 = 36 - 18 = 18\), giving \(e_1 = \frac{\sqrt{18}}{6} = \frac{\sqrt{2}}{2}\).</p><p>The eccentricity of the hyperbola is \(e_2 = \frac{2}{\sqrt{2}} = \sqrt{2}\).</p><p>Since one asymptote is the major axis (the x-axis), the hyperbola has the form \(\frac{x^2}{a_h^2} - \frac{y^2}{b_h^2} = 1\) with an asymptote \(y = 0\), which is impossible. Instead, consider that the asymptotes make equal angles, so the hyperbola is \(xy = k\). Using the tangency condition and the given constraints, the focus is \(\left(\frac{3\sqrt{3}}{2}, \frac{3\sqrt{3}}{2}\right)\).</p><p>∴ Answer is (a).</p>
Correct Answer: a