Ellipse
Confocal Ellipses and Hyperbolas
Grade 11
Question:
<p>For two confocal conics with <span class="math">b' = a</span>, <span class="math">a'e' = b</span>, and <span class="math">4r^2 = 4a^2 + 4b^2</span>, find the eccentricity.</p>
<p>(a) <span class="math">e = \frac{1 + \sqrt{5}}{2}</span></p>
<p>(b) <span class="math">e = \frac{-1 + \sqrt{5}}{2}</span></p>
<p>(c) <span class="math">e = \frac{\sqrt{5} - 1}{2}</span></p>
<p>(d) <span class="math">e = \frac{1}{2}</span></p>
Step-by-Step Solution
Key Concept: Use properties of confocal conics and solve the resulting quadratic equation for eccentricity.
<p><strong>Step 1:</strong> From <span class="math">b' = a</span> and <span class="math">a'e' = b</span>, we have confocal conics.</p><p><strong>Step 2:</strong> The relation <span class="math">b'^2 = a'^2 - a'^2e'^2</span> gives us <span class="math">a^2 = a'^2(1 - e'^2)</span>.</p><p><strong>Step 3:</strong> Solving <span class="math">e^2 + e - 1 = 0</span> using the quadratic formula: <span class="math">e = \frac{-1 \pm \sqrt{5}}{2}</span>.</p><p><strong>Step 4:</strong> Since <span class="math">0 < e < 1</span>, we have <span class="math">e = \frac{\sqrt{5} - 1}{2}</span>.</p><p>∴ Answer is (c).</p>
Correct Answer: C