<p>If the normal to the rectangular hyperbola \(x^2 - y^2 = 4\) at a point P meets the coordinates axes in Q and R and O is the centre of the hyperbola. Then which of the following is correct?</p>
Step-by-Step Solution
Key Concept: Find the equation of the normal at a point P on the hyperbola, then determine where it intersects the coordinate axes (Q and R). Use distance formula to compare OP, PQ, PR, and QR.
<p><strong>Step 1:</strong> Consider a point P on the hyperbola x² - y² = 4. Let P = (2sec θ, 2tan θ) where sec²θ - tan²θ = 1 (standard parametric form).</p><p><strong>Step 2:</strong> Find the equation of tangent at P: Differentiating x² - y² = 4 gives 2x - 2y(dy/dx) = 0, so dy/dx = x/y. At P, slope of tangent = (2sec θ)/(2tan θ) = cos θ/sin θ.</p><p><strong>Step 3:</strong> Slope of normal at P = -sin θ/cos θ. Equation of normal: y - 2tan θ = (-sin θ/cos θ)(x - 2sec θ).</p><p><strong>Step 4:</strong> Find Q (intersection with x-axis, y = 0): 0 - 2tan θ = (-sin θ/cos θ)(x - 2sec θ) → -2sin θ/cos θ = (-sin θ/cos θ)(x - 2sec θ) → x = 4sec θ. So Q = (4sec θ, 0).</p><p><strong>Step 5:</strong> Find R (intersection with y-axis, x = 0): y - 2tan θ = (-sin θ/cos θ)(0 - 2sec θ) → y = 2tan θ + 2sin θ/cos θ · sec θ = 2tan θ + 2tan θ = 4tan θ. So R = (0, 4tan θ).</p><p><strong>Step 6:</strong> Calculate distances:</p><p>PQ² = (2sec θ - 4sec θ)² + (2tan θ - 0)² = 4sec²θ + 4tan²θ = 4(sec²θ + tan²θ)</p><p>PR² = (2sec θ - 0)² + (2tan θ - 4tan θ)² = 4sec²θ + 4tan²θ = 4(sec²θ + tan²θ)</p><p>∴ PQ = PR</p><p><strong>Step 7:</strong> Calculate OP: OP² = (2sec θ)² + (2tan θ)² = 4sec²θ + 4tan²θ = 4(sec²θ + tan²θ)</p><p>So OP = PQ = PR</p><p><strong>Step 8:</strong> Calculate QR: QR² = (4sec θ - 0)² + (0 - 4tan θ)² = 16sec²θ + 16tan²θ = 16(sec²θ + tan²θ)</p><p>∴ QR = 4√(sec²θ + tan²θ) = 2 · 2√(sec²θ + tan²θ) = 2OP</p><p><strong>Verification:</strong> Both PQ = PR and QR = 2OP are verified to be correct.</p><p><strong>∴ Answer:</strong> c,d</p>
Correct Answer: c,d