The equation of the normal drawn to the curve | vedclass

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(D) Given the curve $y^3=4 x^5$. Differentiating both sides with respect to $x$,we get $3 y^2 \frac{d y}{d x} = 20 x^4$. Thus,$\frac{d y}{d x} = \frac

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$3 x+20 y=332$

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(D) Given the curve $y^3=4 x^5$. Differentiating both sides with respect to $x$,we get $3 y^2 \frac{d y}{d x} = 20 x^4$. Thus,$\frac{d y}{d x} = \frac{20 x^4}{3 y^2}$. At the point $(4, 16)$,the slope of the tangent is $m_t = \frac{20(4)^4}{3(16)^2} = \frac{20 	imes 256}{3 	imes 256} = \frac{20}{3}$. The slope of the normal $m_n$ is given by $-\frac{1}{m_t} = -\frac{3}{20}$. The equation of the normal at $(4, 16)$ is $y - 16 = -\frac{3}{20}(x - 4)$. Multiplying by $20$,we get $20y - 320 = -3x + 12$. Rearranging the terms,we get $3x + 20y = 332$.

The equation of the normal drawn to the curve $y^3=4 x^5$ at the point $(4,16)$ is

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