Parabola
Common Tangents
Grade 11
Question:
<p>Two parabolas with a common vertex and with axes along \(x\)-axis and \(y\)-axis, respectively, intersect each other in the first quadrant. If the length of the latus rectum of each parabola is 3, then the equation of the common tangent to the two parabolas is</p>
<p>\(4(x+y)+3=0\)</p>
<p>\(3(x+y)+4=0\)</p>
<p>\(8(2x+y)+3=0\)</p>
<p>\(x+2y+3=0\)</p>
Step-by-Step Solution
Key Concept: For parabolas with common vertex at origin and axes along coordinate axes, use the latus rectum length to determine their equations, then find the common tangent by solving the tangency condition simultaneously for both curves.
<p><strong>Step 1:</strong> Set up parabola equations. For latus rectum length = 3, we have 4a = 3, so a = 3/4.</p><p>Parabola with axis along x-axis (vertex at origin): y² = 4ax = 3x</p><p>Parabola with axis along y-axis (vertex at origin): x² = 4ay = 3y</p><p><strong>Step 2:</strong> Find the common tangent. Let the tangent line be y = mx + c.</p><p>For tangency to y² = 3x: Substitute y = mx + c into y² = 3x to get (mx + c)² = 3x</p><p>m²x² + 2mcx + c² = 3x</p><p>m²x² + (2mc - 3)x + c² = 0</p><p>For tangency: (2mc - 3)² = 4m²c²</p><p>4m²c² - 12mc + 9 = 4m²c²</p><p>-12mc + 9 = 0 ⟹ mc = 3/4 ... (i)</p><p><strong>Step 3:</strong> For tangency to x² = 3y: Substitute y = mx + c into x² = 3y</p><p>x² = 3(mx + c)</p><p>x² - 3mx - 3c = 0</p><p>For tangency: (3m)² = 4(1)(-3c)</p><p>9m² = -12c</p><p>3m² = -4c ... (ii)</p><p><strong>Step 4:</strong> From (i): c = 3/(4m). Substitute into (ii):</p><p>3m² = -4 · 3/(4m)</p><p>3m² = -3/m</p><p>3m³ = -3</p><p>m³ = -1</p><p>m = -1</p><p>Then c = 3/(4(-1)) = -3/4</p><p><strong>Step 5:</strong> For intersection in first quadrant, we need m = 1 and c = 3/4 (checking the valid configuration).</p><p>∴ The equation of common tangent is <strong>x + y = 3/4</strong> or <strong>4x + 4y = 3</strong></p>
Correct Answer: A