Get a normal equation of each line to the curve y = x ^ 3-3x that is parallel to the line 2x+18y-9 = 0The normal line to a graph is the line that...

Here we need to find the normal equation of each line to
the curve  y = x^3 - 3x that is parallel to the line 2x +18y - 9 =
0.

The slope of the tangent to the curve y = x^3 - 3x is
given as y'=3x^2 - 3.

Now for the line 2x + 18y -9
expressing it as y =mx+c where m is the slope and c is the
y-intercept

=> 18y = - 2x + 9=> y = (-1/9)x
+1/2

Therefore the slope is
-1/9.

The slope of the normal therefore is
9.

Now if slope =9

=>
3x^2 - 3 = 9

=> x^2 - 1 =
3

=> x^2 = 4

=>
x = 2 or x = -2

Substituting this in y = 3x^3 -
3x

y = 3*2^3 - 3*2 = 18 for x=
2

and y = -18 for x=-2

So we
need to find lines passing through (2, 18) and (-2, -18) with a slope
-1/9

Now for (2, 18): 18 = (-1/9)*2
+c

=> c = 164/9

Or the
line is y = (-1/9)x + 164/9

=> 9y + x - 164 =
0

And for (-2, -18): -18 = (1/9)*2
+c

=> c = -164/9

Or the
line is y = (-1/9)x - 164/9

=> 9y + x + 164
=0

The required lines are 9y + x - 164 = 0
and 9y + x + 164=0