Question:

<p><strong>Solution:</strong></p><p><strong>Step 1:</strong> Find point on parabola where normal has slope $-1$.</p><p>For $y = x^2 + 1$, slope of tangent is $\frac{dy}{dx} = 2x$.</p><p>Slope of normal is $-\frac{1}{2x}$.</p><p>Setting $-\frac{1}{2x} = -1$: $2x = 1$ ⟹ $x = \frac{1}{2}$</p><p>At $x = \frac{1}{2}$: $y = \frac{1}{4} + 1 = \frac{5}{4}$</p><p><strong>Step 2:</strong> Find equation of normal line through $\left(\frac{1}{2}, \frac{5}{4}\right)$ with slope $-1$:</p><p>$y - \frac{5}{4} = -1\left(x - \frac{1}{2}\right)$</p><p>$y = -x + \frac{1}{2} + \frac{5}{4} = -x + \frac{7}{4}$</p><p><strong>Step 3:</strong> Find intersection points of $y = x^2 + 1$ and $y = -x + \frac{7}{4}$:</p><p>$x^2 + 1 = -x + \frac{7}{4}$</p><p>$x^2 + x - \frac{3}{4} = 0$</p><p>$4x^2 + 4x - 3 = 0$</p><p>$x = \frac{-4 \pm \sqrt{16 + 48}}{8} = \frac{-4 \pm 8}{8}$</p><p>$x = \frac{1}{2}$ or $x = -\frac{3}{2}$</p><p><strong>Step 4:</strong> Calculate enclosed area:</p><p>$A = \int_{-3/2}^{1/2} \left[\left(-x + \frac{7}{4}\right) - (x^2 + 1)\right]dx$</p><p>$= \int_{-3/2}^{1/2} \left(-x^2 - x + \frac{3}{4}\right)dx$</p><p>$= \left[-\frac{x^3}{3} - \frac{x^2}{2} + \frac{3x}{4}\right]_{-3/2}^{1/2}$</p><p>$= \left(-\frac{1}{24} - \frac{1}{8} + \frac{3}{8}\right) - \left(\frac{9}{8} - \frac{9}{8} - \frac{9}{8}\right)$</p><p>$= \frac{4}{3}$</p><p>∴ Answer is (d) $\frac{4}{3}$</p>