<p>The normal to a curve at \(P(x, y)\) meets the \(x\)-axis at \(G\). If the distance of \(G\) from the origin is twice the abscissa of \(P\), then the curve is a/an</p>
Step-by-Step Solution
Key Concept: The normal at point P(x,y) has slope -1/(dy/dx). Setting up the normal line equation and using the condition that it meets x-axis at G where OG = 2x allows us to derive a differential equation that identifies the curve type.
<p><strong>Step 1:</strong> Let the curve be y = f(x). The normal at P(x,y) has slope -1/(dy/dx).</p><p><strong>Step 2:</strong> Equation of normal at P(x,y): Y - y = -1/(dy/dx) · (X - x)</p><p><strong>Step 3:</strong> The normal meets x-axis at G, so put Y = 0: -y = -1/(dy/dx) · (X - x)</p><p>This gives: X = x + y·(dy/dx)</p><p><strong>Step 4:</strong> Given condition: Distance OG = 2x, so |X| = 2x</p><p>Therefore: x + y·(dy/dx) = 2x</p><p>⟹ y·(dy/dx) = x</p><p><strong>Step 5:</strong> Rearranging: y·dy = x·dx</p><p><strong>Step 6:</strong> Integrating both sides: ∫y dy = ∫x dx</p><p>⟹ y²/2 = x²/2 + C</p><p>⟹ y² = x² + 2C</p><p>⟹ y² - x² = constant</p><p><strong>Step 7:</strong> This is the equation of a <strong>rectangular hyperbola</strong> (when constant ≠ 0).</p><p>∴ Answer: A (Ellipse is incorrect; the curve is a rectangular hyperbola)</p>
Correct Answer: A