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

Toolbox:
• 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
$\large\frac{dy}{dx}$$=3x^2-3 \large\frac{dy}{dx_{(x_1,y_1)}}$$=3{x_1}^2-3=9$
=>$3{x_1}^2=12$
$x_1^2=4=>x_1=\pm 2$
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 $(2,2)$
$y-2=-\large\frac{1}{9}$$(x-2) 9y-18=-x+2 x+9y-20=0 Equation of the normal at (-2,-2) y+2=\large\frac{-1}{9}$$(x+2)$
$9y+18=-x-2$
$x+9y+20=0$
The two normals are $x+9y\pm 20=0$