Find the equation of the normal to the curve | vedclass

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(C) Given line is $2x+18y=9$. The slope of this line is $m = -\frac{2}{18} = -\frac{1}{9}$.Given curve is $y=x^3-3x$. Differentiating with respect to

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$x+9y=\pm 20$

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(C) Given line is $2x+18y=9$. The slope of this line is $m = -\frac{2}{18} = -\frac{1}{9}$.Given curve is $y=x^3-3x$. Differentiating with respect to $x$,we get $\frac{dy}{dx} = 3x^2-3$.The slope of the normal is $-\frac{1}{\frac{dy}{dx}} = -\frac{1}{3x^2-3}$.Since the normal is parallel to the given line,their slopes are equal:$-\frac{1}{9} = -\frac{1}{3(x^2-1)} \Rightarrow 3(x^2-1) = 9 \Rightarrow x^2-1 = 3 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2$.Case $1$: If $x=2$,then $y = (2)^3 - 3(2) = 8-6 = 2$. The point is $(2, 2)$.The equation of the normal is $y-2 = -\frac{1}{9}(x-2) \Rightarrow 9y-18 = -x+2 \Rightarrow x+9y = 20$.Case $2$: If $x=-2$,then $y = (-2)^3 - 3(-2) = -8+6 = -2$. The point is $(-2, -2)$.The equation of the normal is $y-(-2) = -\frac{1}{9}(x-(-2)) \Rightarrow 9(y+2) = -(x+2) \Rightarrow 9y+18 = -x-2 \Rightarrow x+9y = -20$.Combining both cases,the required equations are $x+9y = \pm 20$.

Find the equation of the normal to the curve $y=x^3-3x$ which is parallel to the line $2x+18y=9$.

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