<p>Tangent at P to rectangular hyperbola <span class="inline-math">\(xy = 2\)</span> meets coordinate axes at A and B, then area of triangle OAB (where O is origin) is:</p>
Step-by-Step Solution
Key Concept: For a rectangular hyperbola xy = c², the tangent at any point has a special property: it makes equal intercepts on the coordinate axes, and the area of the triangle formed by the tangent and the axes is constant (equal to 2c²).
<p><strong>Step 1:</strong> Write the equation of tangent at point P(t, 2/t) on the hyperbola xy = 2.</p><p>For rectangular hyperbola xy = c², the tangent at point (ct, c/t) is given by: tx + y = 2c.</p><p>Here c² = 2, so c = √2. At point P(t, 2/t), the tangent equation is: <strong>tx + y = 2√2</strong>.</p><p><strong>Step 2:</strong> Find intercepts A and B on the coordinate axes.</p><p>For x-intercept (point A), set y = 0: tx = 2√2, so x = 2√2/t. Thus A = (2√2/t, 0).</p><p>For y-intercept (point B), set x = 0: y = 2√2. Thus B = (0, 2√2).</p><p><strong>Step 3:</strong> Calculate the area of triangle OAB.</p><p>The area of triangle with vertices O(0,0), A(2√2/t, 0), and B(0, 2√2) is:</p><p>Area = (1/2) × base × height = (1/2) × |2√2/t| × |2√2|</p><p>Area = (1/2) × (2√2/t) × 2√2 = (1/2) × (8/t) × t = (1/2) × 8 = <strong>4</strong></p><p>Note: The parameter t cancels out, proving the area is constant for any point on the hyperbola.</p><p><strong>∴ Answer:</strong> 4</p>
Correct Answer: 4