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 is perpendicular to the line tangent.

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neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

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The normal equation to the curve at any point is perpendicular to the tangent at any point on the curve y = x^3-3x. The slope of the tangent to the curve is dy/dx. Therefore the slope of the normal to the curve is -1/(dy/dx).

So we find the dy/dx firsy.

y = x^3 -3x.

dy/dx = d/dx{x^3-3x) = 3x^2-3.

Therefore the slope of the nomal at any point = -1/(dy/dx) = -1/(3x^2-3). But this slope  -1/(3x-3) is  equal to the slope of the line 2x+18y-9 = 0, the normal is given to be parallel to the line 2x+18t-9 - 0 The slope the line ax+by+c = 0 is -x/b. So the slope of 2x+18y-9 = 0 is  -2/18.

-1/(3x-3) = -2/18

18 = 2(3x-3) = 6(x-1)

Divide by 6:

3 = x-1

x = 4

Therefore the the y coordinate of the normal to the curve is got by putting x = 4 in y = x^3-3x. So y = 4^3-3*4 =  52

Therefore the normal at (4,52) is a line whose slope is -2/18 = -1/9.

The equation of the line through (x1, y1) with a slope m is given by:

y-y1 = m (x-x1).

Therefore the equation of the normal at (4,52) with slope -1/9 is given bY:

y -52 = -(1/9)(x-4)

(y -52)9 =(4-x)

x+9y - 52*9-4 = 0

x+9y - 472 = 0.

william1941's profile pic

william1941 | College Teacher | (Level 3) Valedictorian

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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

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