Ellipse
Standard form of Ellipse
Grade 11
Question:
<p>\(\dfrac{x^2}{r^2 - r - 6} + \dfrac{y^2}{r^2 - 6r + 5} = 1\) will represent the ellipse, if \(r\) lies in the interval</p>
<p>(a) \((-\infty, -2)\)</p>
<p>(b) \((3, \infty)\)</p>
<p>(c) \((5, \infty)\)</p>
<p>(d) \((1, \infty)\)</p>
Step-by-Step Solution
Key Concept: For the equation to represent an ellipse, both denominators must be positive and unequal: a² > 0, b² > 0, and a² ≠ b². This requires finding where both quadratic expressions are positive and where they differ.
<p><strong>Step 1:</strong> For an ellipse, both denominators must be positive and unequal.</p><p><strong>Step 2:</strong> Factor the denominators:</p><p>• r² - r - 6 = (r - 3)(r + 2) > 0 ⟹ r < -2 or r > 3</p><p>• r² - 6r + 5 = (r - 1)(r - 5) > 0 ⟹ r < 1 or r > 5</p><p><strong>Step 3:</strong> Find intersection of both conditions:</p><p>From r < -2 or r > 3 AND r < 1 or r > 5</p><p>• For r < -2: r < -2 automatically satisfies r < 1, so r < -2 works</p><p>• For r > 3: we need r > 5 (to satisfy r > 5 part), so r > 5 works</p><p>Combined: r ∈ (-∞, -2) ∪ (5, ∞)</p><p><strong>Step 4:</strong> Check a² ≠ b² condition:</p><p>r² - r - 6 ≠ r² - 6r + 5</p><p>-r - 6 ≠ -6r + 5</p><p>5r ≠ 11 ⟹ r ≠ 11/5 = 2.2</p><p>This is already excluded from our solution, so no additional restriction.</p><p>∴ Answer: r ∈ (-∞, -2) ∪ (5, ∞)</p>
Correct Answer: C