<p>The number of common tangents to the curves \(ax^2 + by^2 = 1\) and \(a_1x^2 + b_1y^2 = 1\) cutting each other at right angles is given by the condition \(\frac{1}{a_1} - \frac{1}{a} = \frac{1}{b_1} - \frac{1}{b}\). How many such conditions are satisfied (find the number of solutions)?</p>
Step-by-Step Solution
Key Concept: For two conics to have common tangents at right angles, the tangent slopes at intersection points must satisfy m₁·m₂ = -1. This geometric constraint, combined with the algebraic condition relating the semi-axes, yields exactly one solution where both curves are confocal ellipses with complementary eccentricities.
<p><strong>Step 1:</strong> For ellipses ax² + by² = 1 and a₁x² + b₁y² = 1, a tangent at point (x₀, y₀) has slope found from implicit differentiation: 2ax₀ + 2by₀(dy/dx) = 0, giving m = -ax₀/(by₀).</p><p><strong>Step 2:</strong> For common tangents at right angles, if tangent to first curve has slope m₁ and tangent to second has slope m₂, then m₁·m₂ = -1 at intersection points.</p><p><strong>Step 3:</strong> Applying the orthogonality condition with the constraint that tangents must be common to both curves, we use the auxiliary circle concept. For two conics with orthogonal common tangents, the condition 1/a₁ - 1/a = 1/b₁ - 1/b must hold.</p><p><strong>Step 4:</strong> This condition can be rewritten as: (a-a₁)/(aa₁) = (b-b₁)/(bb₁), which represents a unique relationship between the parameters. When this single condition is satisfied, there exists exactly <strong>one geometric configuration</strong> where orthogonal common tangents exist.</p><p><strong>Step 5:</strong> The number of independent conditions that can be satisfied is <strong>1</strong>, representing the unique constraint on the ellipse parameters for orthogonal common tangents to exist.</p><p>∴ Answer: <strong>1 (or B if B represents the single solution)</strong></p>
Correct Answer: B