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

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

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$4 x+3 y=7 a$

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(C) Given curve is $y^4=a x^3$.On differentiating with respect to $x$,we get:$4 y^3 \frac{d y}{d x} = 3 a x^2$.At the point $(a, a)$,the slope of the tangent is:$\left(\frac{d y}{d x}ight)_{(a, a)} = \frac{3 a(a)^2}{4(a)^3} = \frac{3 a^3}{4 a^3} = \frac{3}{4}$.The slope of the normal is the negative reciprocal of the tangent slope:$m_{	ext{normal}} = -\frac{1}{3/4} = -\frac{4}{3}$.The equation of the normal at point $(a, a)$ is given by $y - y_1 = m(x - x_1)$:$y - a = -\frac{4}{3}(x - a)$.Multiplying by $3$: $3y - 3a = -4x + 4a$.Rearranging the terms: $4x + 3y = 7a$.

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

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