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(D) Given the curve $xy^2 + x^2y = 12$. Differentiating with respect to $x$: $y^2 + 2xy \frac{dy}{dx} + 2xy + x^2 \frac{dy}{dx} = 0$. At the point $(1

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$\frac{274}{35}$

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(D) Given the curve $xy^2 + x^2y = 12$. Differentiating with respect to $x$: $y^2 + 2xy \frac{dy}{dx} + 2xy + x^2 \frac{dy}{dx} = 0$. At the point $(1, 3)$, we have $3^2 + 2(1)(3) \frac{dy}{dx} + 2(1)(3) + (1)^2 \frac{dy}{dx} = 0$. $9 + 6 \frac{dy}{dx} + 6 + \frac{dy}{dx} = 0 \implies 7 \frac{dy}{dx} = -15 \implies \frac{dy}{dx} = -\frac{15}{7}$. The equation of the tangent at $(1, 3)$ is $y - 3 = -\frac{15}{7}(x - 1)$. For $T$, set $y = 0$: $-3 = -\frac{15}{7}(x - 1) \implies 21 = 15(x - 1) \implies x - 1 = \frac{21}{15} = \frac{7}{5} \implies x = 1 + \frac{7}{5} = \frac{12}{5}$. So $T = (\frac{12}{5}, 0)$. The equation of the normal at $(1, 3)$ is $y - 3 = \frac{7}{15}(x - 1)$. For $N$, set $y = 0$: $-3 = \frac{7}{15}(x - 1) \implies -45 = 7(x - 1) \implies x - 1 = -\frac{45}{7} \implies x = 1 - \frac{45}{7} = -\frac{38}{7}$. So $N = (-\frac{38}{7}, 0)$. $TN = |\frac{12}{5} - (-\frac{38}{7})| = |\frac{12}{5} + \frac{38}{7}| = |\frac{84 + 190}{35}| = \frac{274}{35}$.

If the tangent and the normal drawn to the curve $xy^2 + x^2y = 12$ at the point $(1, 3)$ meet the $X$-axis in $T$ and $N$ respectively, then $TN =$

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