<p>Which of the following is/are true about the ellipse \(x^2 + 4y^2 - 2x - 16y + 13 = 0\)?</p><p>(a) The latus rectum of the ellipse is 1.</p><p>(b) Distance between foci of the ellipse is \(4\sqrt{3}\).</p><p>(c) Sum of the focal distances of a point \(P(x, y)\) on the ellipse is 4.</p><p>(d) \(y = 3\) meets the tangents drawn at the vertices of the ellipse at points \(P\) and \(Q\), then \(PQ\) subtends a right angle at any of its foci.</p>
<p>(a) The latus rectum of the ellipse is 1.</p>
<p>(b) Distance between foci of the ellipse is \(4\sqrt{3}\).</p>
<p>(c) Sum of the focal distances of a point \(P(x, y)\) on the ellipse is 4.</p>
<p>(d) \(y = 3\) meets the tangents drawn at the vertices of the ellipse at points \(P\) and \(Q\), then \(PQ\) subtends a right angle at any of its foci.</p>
Step-by-Step Solution
Key Concept: First convert the ellipse equation to standard form by completing the square, then identify a, b, c to verify each property systematically. The sum of focal distances equals 2a regardless of point position on the ellipse.
<p><strong>Step 1: Complete the square</strong></p><p>x² + 4y² - 2x - 16y + 13 = 0</p><p>(x² - 2x + 1) + 4(y² - 4y + 4) + 13 - 1 - 16 = 0</p><p>(x - 1)² + 4(y - 2)² = 4</p><p><strong>Step 2: Convert to standard form</strong></p><p>\frac{(x-1)^2}{4} + \frac{(y-2)^2}{1} = 1</p><p>Center: (1, 2), a² = 4, b² = 1, so a = 2, b = 1</p><p><strong>Step 3: Calculate c</strong></p><p>c² = a² - b² = 4 - 1 = 3, so c = √3</p><p><strong>Check (a): Latus rectum = 2b²/a = 2(1)/2 = 1 ✓ TRUE</strong></p><p><strong>Check (b): Distance between foci = 2c = 2√3 ≠ 4√3 ✗ FALSE</strong></p><p><strong>Check (c): Sum of focal distances = 2a = 4 for any point on ellipse ✓ TRUE</strong></p><p><strong>Check (d): Vertices on major axis at (3,2) and (-1,2). Tangents at vertices are x = 3 and x = -1 (vertical lines). These don't meet y = 3. Check minor axis vertices (1,3) and (1,1).</strong></p><p><strong>Tangent at (1,3): y - 3 = 0 or y = 3 (horizontal). Tangent at (1,1): y = 1.</strong></p><p><strong>The line y = 3 is itself tangent at (1,3). For the tangent at (1,1): it's horizontal at y = 1.</strong></p><p><strong>Reinterpret: tangents at endpoints of major axis (±3,2) relative to center give x = ±3 (shifted). Line y = 3 meets these at P(3,3), Q(-1,3).</strong></p><p><strong>Foci at (1±√3, 2). Check angle: slope from F₁(1-√3,2) to P(3,3) and to Q(-1,3) gives angle = 90° ✓ TRUE</strong></p><p><strong>∴ Answer: A, C, D</strong>
Correct Answer: A,C,D