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(B) Given the curve $y^2 = x^3$ at $x = 8$.Substituting $x = 8$ into the equation: $y^2 = 8^3 = 512$,so $y = \pm \sqrt{512} = \pm 16\sqrt{2}$.The poin

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$x \pm 3\sqrt{2} y = 104$

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(B) Given the curve $y^2 = x^3$ at $x = 8$.Substituting $x = 8$ into the equation: $y^2 = 8^3 = 512$,so $y = \pm \sqrt{512} = \pm 16\sqrt{2}$.The points are $(8, 16\sqrt{2})$ and $(8, -16\sqrt{2})$.Differentiating $y^2 = x^3$ with respect to $x$: $2y \frac{dy}{dx} = 3x^2$,which gives $\frac{dy}{dx} = \frac{3x^2}{2y}$.At $(8, \pm 16\sqrt{2})$,the slope of the tangent $m_t = \frac{3(8)^2}{2(\pm 16\sqrt{2})} = \frac{3 	imes 64}{\pm 32\sqrt{2}} = \pm 3\sqrt{2}$.The slope of the normal $m_n = -\frac{1}{m_t} = \mp \frac{1}{3\sqrt{2}}$.The equation of the normal is $(y - y_1) = m_n(x - x_1)$.For $(8, \pm 16\sqrt{2})$: $(y \mp 16\sqrt{2}) = \mp \frac{1}{3\sqrt{2}}(x - 8)$.Multiplying by $\mp 3\sqrt{2}$: $\mp 3\sqrt{2}y + 96 = x - 8$.Rearranging gives $x \pm 3\sqrt{2}y = 104$.

What is the equation of the normal to the curve $y^2 = x^3$ at the point where the $x$-coordinate is $8$?

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