The equation of the normal to the curve y^4=a | vedclass

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(C) Given curve is $y^4=ax^3$. On differentiating with respect to $x$, we get: $4y^3 \frac{dy}{dx} = 3ax^2$. At the point $(a, a)$, the slope of the t

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$4x+3y=7a$

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(C) Given curve is $y^4=ax^3$. On differentiating with respect to $x$, we get: $4y^3 \frac{dy}{dx} = 3ax^2$. At the point $(a, a)$, the slope of the tangent is: $\frac{dy}{dx} = \frac{3a(a)^2}{4(a)^3} = \frac{3a^3}{4a^3} = \frac{3}{4}$. The slope of the normal is the negative reciprocal of the slope of the tangent: $m_{	ext{normal}} = -\frac{1}{3/4} = -\frac{4}{3}$. The equation of the normal at $(a, a)$ is given by: $y - a = -\frac{4}{3}(x - a)$. Multiplying by $3$: $3y - 3a = -4x + 4a$. Rearranging the terms: $4x + 3y = 7a$.

The equation of the normal to the curve $y^4=ax^3$ at $(a, a)$ is

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