Parabola
Common Tangents
Grade 11
Question:
<p>Equation of a common tangent to the circle, \(x^2 + y^2 - 6x = 0\) and the parabola, \(y^2 = 4x\), is:</p>
<p>\(2\sqrt{3}y = 12x + 1\)</p>
<p>\(\sqrt{3}y = x + 3\)</p>
<p>\(2\sqrt{3}y = -x - 12\)</p>
<p>\(\sqrt{3}y = 3x + 1\)</p>
Step-by-Step Solution
Key Concept: A common tangent must satisfy the tangency condition for both curves simultaneously. For the parabola y² = 4x, use parametric form or the condition that discriminant = 0 when the tangent line intersects it. For the circle, use the perpendicular distance from center equals radius.
<p><strong>Step 1:</strong> Rewrite the circle: x² + y² - 6x = 0 → (x-3)² + y² = 9. Center C(3, 0), radius r = 3.</p><p><strong>Step 2:</strong> For parabola y² = 4x, a tangent has form y = mx + 1/m (standard form, where a = 1).</p><p><strong>Step 3:</strong> Apply tangency condition for circle: Distance from C(3,0) to line mx - y + 1/m = 0 equals 3.</p><p>$$\frac{|3m - 0 + 1/m|}{\sqrt{m^2 + 1}} = 3$$</p><p><strong>Step 4:</strong> Simplify: |3m + 1/m|² = 9(m² + 1)</p><p>$$9m^2 + 6 + \frac{1}{m^2} = 9m^2 + 9$$</p><p>$$\frac{1}{m^2} = 3$$</p><p>$$m = \pm\frac{1}{\sqrt{3}}$$</p><p><strong>Step 5:</strong> Substituting back: y = ±1/√3 · x ± √3</p><p>Simplifying: <strong>x - √3y + 3 = 0</strong> or <strong>x + √3y + 3 = 0</strong></p><p>∴ Answer: B</p>
Correct Answer: B