<p>Find the equations to the common tangents to the circle \(x^2 + y^2 = 2a^2\) and the parabola \(y^2 = 8ax\).</p>
<p>\(y = x + 2a\) and \(y = -(x + 2a)\)</p>
<p>\(y = x - 2a\) and \(y = -(x - 2a)\)</p>
<p>\(y = 2x + a\) and \(y = -(2x + a)\)</p>
<p>\(y = x + a\) and \(y = -(x + a)\)</p>
Step-by-Step Solution
Key Concept: A common tangent to both curves must satisfy the tangency condition for the circle (distance from center = radius) and for the parabola (discriminant = 0). Use the parametric form of parabola tangent y = mx + 2a/m and apply the circle tangency condition.
<p><strong>Step 1:</strong> For parabola y² = 8ax, the tangent at parameter t is: y = (x/2t) + 2at, or in slope form: y = mx + 2a/m where m is the slope.</p><p><strong>Step 2:</strong> For this line y = mx + 2a/m to be tangent to circle x² + y² = 2a², the distance from origin must equal radius √(2a²) = a√2.</p><p><strong>Step 3:</strong> Distance from (0,0) to line mx - y + 2a/m = 0 is: |2a/m|/√(m² + 1) = a√2</p><p><strong>Step 4:</strong> Squaring: 4a²/m² = 2a²(m² + 1), which gives 4a²/m² = 2a²m² + 2a²</p><p><strong>Step 5:</strong> Simplifying: 4/m² = 2m² + 2, so 4 = 2m⁴ + 2m², giving m⁴ + m² - 2 = 0</p><p><strong>Step 6:</strong> Let u = m²: u² + u - 2 = 0, so (u + 2)(u - 1) = 0. Since u = m² ≥ 0, we get m² = 1, thus m = ±1</p><p><strong>Step 7:</strong> For m = 1: y = x + 2a. For m = -1: y = -x - 2a</p><p>∴ <strong>Answer:</strong> y = x + 2a and y = -x - 2a (or x - y + 2a = 0 and x + y + 2a = 0)</p>
Correct Answer: A