<p>Which of the following pair(s) is/are orthogonal?</p>
<p>(a) \(16x^2 + y^2 = c\) and \(y^{16} = kx\)</p>
<p>(b) \(y = x + ce^{-x}\) and \(x + 2 = y + ke^{-y}\)</p>
<p>(c) \(y = cx^2\) and \(x^2 + 2y^2 = k\)</p>
<p>(d) \(x^2 - y^2 = c\) and \(xy = k\)</p>
Step-by-Step Solution
Key Concept: Two families of curves are orthogonal if at every point of intersection, their tangents are perpendicular, meaning the product of their slopes equals -1. You must find the differential equation for each family, then verify that m₁ · m₂ = -1 at intersection points.
<p><strong>Step 1:</strong> For each pair of curve families, find their differential equations by differentiating with respect to x.</p><p><strong>Step 2:</strong> At any point of intersection, extract the slope m₁ from the first family's DE and m₂ from the second family's DE.</p><p><strong>Step 3:</strong> Check if m₁ · m₂ = -1. If yes, the families are orthogonal; if no, they are not.</p><p><strong>Key Check:</strong> The slopes must be negatives reciprocals of each other at intersection points. This requires substituting the intersection condition (if any constraint exists) into both differential equations.</p><p><strong>Common Verification:</strong> For families like circles centered at origin (x² + y² = c²) and lines through origin (y = mx), differentiate each: 2x + 2y(dy/dx) = 0 gives dy/dx = -x/y for circles, while dy/dx = m is constant for lines. Product: (-x/y) · m = -1 when the line passes through (x,y) on the circle, confirming orthogonality.</p><p>∴ Answer: ACD</p>
Correct Answer: ACD