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Find the equations of normal to $y=x^{3}-3x$ that is parallel to $2x+18y-9=0.$

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  • If $y=f(x)$ then $\large\frac{dy}{dx}$$=f'(x)$ is the rate of change of $y$ w.r.t $x$
  • $\large\frac{dy}{dx_{(x_1,y_1)}}$ is the slope of the tangent to the curve at the point $(x_1,y_1) $ on the curve. It is the slope of the curve at that point.
  • The normal at a point $(x_1,y_1)$ on $y=f(x)$ is perpendicular to the tangent at $(x_1,y_1)$
Step 1:
Let $(x_1,y_1)$ be the point at which the normal to $y=x^3-3x$ is $||$ to $2x+18 y-9=0$
The slope of the normal at $(x_2 ,y_1)=\large \frac{-2}{18}=\frac{-1}{9}$$=m$
Therefore the slope of the tangent =9
Step 2:
Slope of the tangent $=\large\frac{dy}{dx _(x_1y_1)}$
Differentiate along $y= x^3-3x$ with respect to x
$x_1^2=4=>x_1=\pm 2$
Step 3:
When $ x_1=2\qquad y_1=8-6=2$ the point is $(2,2)$
When $ x_1=-2\qquad y_1=-8+6=-2$ the point is $(-2,-2)$
Step 4:
Equation of the normal at $(2,2)$
Equation of the normal at $(-2,-2)$
The two normals are $x+9y\pm 20=0$
answered Jul 24, 2013 by meena.p

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