<p>If the tangents drawn to the hyperbola \(4y^2 = x^2 + 1\) intersect the co-ordinate axes at the distinct points \(A\) and \(B\), then the locus of the midpoint of \(AB\) is</p>
<p>\(x^2 - 4y^2 + 16x^2y^2 = 0\)</p>
<p>\(x^2 - 4y^2 - 16x^2y^2 = 0\)</p>
<p>\(4x^2 - y^2 + 16x^2y^2 = 0\)</p>
<p>\(4x^2 - y^2 - 16x^2y^2 = 0\)</p>
Step-by-Step Solution
Key Concept: Rewrite the hyperbola in standard form, find the general tangent equation, determine where it intersects the coordinate axes (points A and B), then find the locus of the midpoint of AB by eliminating the parameter.
<p><strong>Step 1:</strong> Rewrite the hyperbola in standard form: 4y² = x² + 1 → y²/(1/4) - x²/1 = 1</p><p>Here a² = 1/4, b² = 1, so a = 1/2, b = 1</p><p><strong>Step 2:</strong> The tangent to hyperbola y²/a² - x²/b² = 1 is given by: y = mx ± √(a²m² - b²)</p><p>For our hyperbola: y = mx ± √(m²/4 - 1)</p><p><strong>Step 3:</strong> Find x-intercept (point A on x-axis, y=0): 0 = mx ± √(m²/4 - 1) → x = ∓√(m²/4 - 1)/m</p><p>So A = (-√(m²/4 - 1)/m, 0) or (√(m²/4 - 1)/m, 0)</p><p><strong>Step 4:</strong> Find y-intercept (point B on y-axis, x=0): y = ± √(m²/4 - 1)</p><p>So B = (0, √(m²/4 - 1)) or (0, -√(m²/4 - 1))</p><p><strong>Step 5:</strong> Midpoint M of AB: Let M = (h, k) where 2h = ±√(m²/4 - 1)/m and 2k = ±√(m²/4 - 1)</p><p>From these: 2h·m = ±√(m²/4 - 1) and 2k = ±√(m²/4 - 1)</p><p>Therefore: 2hm = 2k → k = hm</p><p><strong>Step 6:</strong> From 2k = √(m²/4 - 1): 4k² = m²/4 - 1 → m² = 16k² + 4</p><p>Substitute m = k/h: (k/h)² = 16k² + 4 → k² = h²(16k² + 4) → k² = 16h²k² + 4h²</p><p>Rearranging: k²(1 - 16h²) = 4h² → x² - 16x²y² = 4x² or <strong>16x²y² = x² - 4x²</strong></p><p>The locus is: <strong>x² - 4y² = 4x²</strong> or equivalently <strong>y² = x²/4 - 1</strong> or <strong>4y² - x² = -4</strong></p><p>∴ Answer: D</p>
Correct Answer: D