What is the equation of the normal to the cur | vedclass

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(B) Given the curve equation: $y^2 = 16x$. Differentiating with respect to $x$: $2y \frac{dy}{dx} = 16$,which gives $\frac{dy}{dx} = \frac{8}{y}$. At

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$x + 2y = 9$

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(B) Given the curve equation: $y^2 = 16x$. Differentiating with respect to $x$: $2y \frac{dy}{dx} = 16$,which gives $\frac{dy}{dx} = \frac{8}{y}$. At the point $(1, 4)$,the slope of the tangent $m_t = \frac{8}{4} = 2$. The slope of the normal $m_n = -\frac{1}{m_t} = -\frac{1}{2}$. The equation of the normal at $(1, 4)$ is given by $y - y_1 = m_n(x - x_1)$. Substituting the values: $y - 4 = -\frac{1}{2}(x - 1)$. Multiplying by $2$: $2y - 8 = -x + 1$,which simplifies to $x + 2y = 9$.

What is the equation of the normal to the curve $y^2 = 16x$ at the point $(1, 4)$?

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