Question:

<p><strong>Solution:</strong> First, find the point where the curve intersects the Y-axis by setting $x = 0$:</p><p>$$y(0-2)(0-3) = 0 + 6$$</p><p>$$y \cdot (-2) \cdot (-3) = 6$$</p><p>$$6y = 6$$</p><p>$$y = 1$$</p><p>So the point of intersection with Y-axis is $(0, 1)$.</p><p>Now find $\frac{dy}{dx}$ from $y(x-2)(x-3) = x + 6$:</p><p>$$y[(x-2)(x-3)]' + (x-2)(x-3)y' = 1$$</p><p>$$y[(x-3) + (x-2)] + (x-2)(x-3)y' = 1$$</p><p>$$y(2x-5) + (x-2)(x-3)y' = 1$$</p><p>At $(0, 1)$:</p><p>$$1(-5) + (-2)(-3)y' = 1$$</p><p>$$-5 + 6y' = 1$$</p><p>$$y' = 1$$</p><p>The slope of the normal is $-1$. Equation of normal at $(0, 1)$:</p><p>$$y - 1 = -1(x - 0)$$</p><p>$$y = -x + 1$$</p><p>Check which point lies on this line: $\left(\frac{1}{2}, \frac{1}{2}\right)$: $\frac{1}{2} = -\frac{1}{2} + 1 = \frac{1}{2}$ ✓</p><p>∴ Answer is (c).</p>