Ellipse
Common Tangent with Hyperbola
Grade 11
Question:
<p>For the ellipse \(\frac{x^2}{36} + \frac{y^2}{18} = 1\) and the hyperbola described above, find the equation of the common tangent in the first quadrant.</p>
<p>(a) \(\frac{x}{2} + y = 3\)</p>
<p>(b) \(\frac{x}{2} + y = 6\)</p>
<p>(c) \(x + y\sqrt{2} = 3\sqrt{2}\)</p>
<p>(d) \(x + y\sqrt{2} = 6\)</p>
Step-by-Step Solution
Key Concept: Apply the tangency condition simultaneously to both the ellipse and hyperbola equations to find the common tangent.
<p><strong>Solution:</strong> A common tangent to both curves must satisfy the tangency conditions for both the ellipse and hyperbola. For the ellipse \(\frac{x^2}{36} + \frac{y^2}{18} = 1\), the tangent of the form \(y = mx + c\) satisfies \(c^2 = 36m^2 + 18\). For the hyperbola, applying the tangency condition with the eccentricity relationship determined earlier, we find that \(\frac{x}{2} + y = 3\) satisfies both conditions in the first quadrant.</p><p>∴ Answer is (a).</p>
Correct Answer: a