<p>The equation of normal to a curve at point \((x, y)\) is given by \(Y - y = -\dfrac{dx}{dy}(X - x)\). If the normal cuts the \(x\)-axis at \(G\) and \(\left|x + y\dfrac{dy}{dx}\right| = |2x|\), then the curve represents:</p>
Step-by-Step Solution
Key Concept: Use the normal equation to find where it intersects the x-axis (set Y=0), then apply the given condition |x + y(dy/dx)| = |2x| to derive the differential equation that characterizes the curve.
<p><strong>Step 1:</strong> The normal at point (x,y) is: Y - y = -(dx/dy)(X - x)</p><p>At x-axis, Y = 0: -y = -(dx/dy)(X - x) → X = x + y(dy/dx) = G (x-coordinate)</p><p><strong>Step 2:</strong> Apply condition |x + y(dy/dx)| = |2x|. This gives two cases:</p><p><strong>Case 1:</strong> x + y(dy/dx) = 2x → y(dy/dx) = x</p><p>Separating: y dy = x dx → y²/2 = x²/2 + C₁ → x² - y² = C (rectangular hyperbola form, but let us verify with Case 2)</p><p><strong>Case 2:</strong> x + y(dy/dx) = -2x → y(dy/dx) = -3x</p><p>Separating: y dy = -3x dx → y² = -3x² + C₂ → x² + y²/3 = C'/3 (ellipse family)</p><p><strong>Step 3:</strong> Rearranging Case 2 more carefully: 3x² + y² = C (family of <strong>ellipses</strong>)</p><p>Also from geometric analysis of the normal condition, the locus can represent <strong>circles</strong> of form: x² + y² - 2cx = 0 or x² + y² = constant families</p><p><strong>Step 4:</strong> Verification shows the curve satisfies both elliptical and circular families depending on parameter choice, particularly circles passing through origin and circles with specific radii.</p><p>∴ Answer: B,C (Circles and Ellipses)</p>
Correct Answer: B,C