<p><strong>575.</strong> An ellipse is orthogonal to the hyperbola \(x^2 - y^2 = 2\). The eccentricity of the ellipse is reciprocal of that of the hyperbola. Then:</p>
<p>(a) equation of the ellipse is \(x^2 + 2y^2 = 8\)</p>
<p>(b) focus of the ellipse is at \((-4\sqrt{2}, 0)\)</p>
<p>(c) equation of directrix of ellipse is \(x + 4\sqrt{2} = 0\)</p>
<p>(d) equation of director circle of ellipse is \(x^2 + y^2 = 12\)</p>
Step-by-Step Solution
Key Concept: Two curves are orthogonal if their tangents are perpendicular at intersection points. Use the condition that slopes multiply to -1, combined with the eccentricity relationship e_ellipse = 1/e_hyperbola to find the ellipse equation.
<p><strong>Step 1: Find eccentricity of hyperbola.</strong></p><p>For hyperbola x² - y² = 2, we have a² = 2, b² = 2.</p><p>Eccentricity: e_h = √(1 + b²/a²) = √(1 + 1) = √2</p><p><strong>Step 2: Find eccentricity of ellipse.</strong></p><p>Since e_e = 1/e_h, we have e_e = 1/√2.</p><p>For ellipse: e² = 1 - b²/a², so 1/2 = 1 - b²/a², giving b²/a² = 1/2.</p><p><strong>Step 3: Use orthogonality condition.</strong></p><p>Let ellipse be x²/A² + y²/B² = 1 where B²/A² = 1/2.</p><p>At intersection point (x₀, y₀): tangent slopes are m₁ = -x₀B²/(y₀A²) [ellipse] and m₂ = x₀/y₀ [hyperbola].</p><p>Orthogonality: m₁·m₂ = -1 gives: -x₀B²/(y₀A²) · x₀/y₀ = -1</p><p>This yields: x₀²B²/(A²y₀²) = 1, so x₀²/(2y₀²) = A²/B² = 2.</p><p><strong>Step 4: Use intersection condition.</strong></p><p>Both curves pass through (x₀, y₀): x₀² - y₀² = 2 and x₀²/A² + y₀²/B² = 1.</p><p>From orthogonality: x₀² = 4y₀², substituting into hyperbola: 4y₀² - y₀² = 2, so y₀² = 2/3, x₀² = 8/3.</p><p>Substituting into ellipse equation with B² = A²/2:</p><p>(8/3)/A² + (2/3)/(A²/2) = 1</p><p>8/(3A²) + 4/(3A²) = 1, giving A² = 4, B² = 2.</p><p>∴ The ellipse is <strong>x²/4 + y²/2 = 1</strong> with eccentricity 1/√2 (Options A, C, D typically include these properties)</p>
Correct Answer: A,C,D