Question:

<p><strong>Solution:</strong> First verify that $(1, 1)$ lies on the curve:</p><p>$$1^2 + 2(1)(1) - 3(1)^2 = 1 + 2 - 3 = 0$$ ✓</p><p>Find $\frac{dy}{dx}$ from $x^2 + 2xy - 3y^2 = 0$:</p><p>$$2x + 2y + 2x\frac{dy}{dx} - 6y\frac{dy}{dx} = 0$$</p><p>$$2x + 2y + (2x - 6y)\frac{dy}{dx} = 0$$</p><p>$$\frac{dy}{dx} = \frac{-(2x + 2y)}{2x - 6y} = \frac{-2(x + y)}{2(x - 3y)}$$</p><p>At $(1, 1)$:</p><p>$$\frac{dy}{dx} = \frac{-2(1 + 1)}{2(1 - 3)} = \frac{-4}{-4} = 1$$</p><p>Slope of normal $= -1$. Equation of normal:</p><p>$$y - 1 = -1(x - 1)$$</p><p>$$y = -x + 2$$</p><p>To find other intersections, substitute $y = -x + 2$ into $x^2 + 2xy - 3y^2 = 0$:</p><p>$$x^2 + 2x(-x + 2) - 3(-x + 2)^2 = 0$$</p><p>$$x^2 - 2x^2 + 4x - 3(x^2 - 4x + 4) = 0$$</p><p>$$-x^2 + 4x - 3x^2 + 12x - 12 = 0$$</p><p>$$-4x^2 + 16x - 12 = 0$$</p><p>$$x^2 - 4x + 3 = 0$$</p><p>$$(x - 1)(x - 3) = 0$$</p><p>$$x = 1 \text{ or } x = 3$$</p><p>When $x = 3$: $y = -3 + 2 = -1$, giving point $(3, -1)$ which is in the fourth quadrant.</p><p>∴ Answer is (d).</p>