Find an equation of the line that is both normal to the given parabola and also parallel to the given line. y = 6 x 2 - 8 x + 9x - 8 y = 9 y=?

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giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

To determine the equation of the line, we'll impose the conditions:

1) The equation of the line is perpendicular to the given parabola if and only if the product of the slope of the line and the derivative of the quadratic function is -1.

2) The line is parallel to the line x-8y = 9 if and only if their slopes are equal.

We'll note the slope of the line that has to be determined as m.

1) We'll calculate the first derivative of y = 6x^2-8x+9

f'(x)=12x-8

The slope of f'(x)=12

2) We'll put x-8y = 9 in the standard form:

y = mx+n

We'll isolate y to the left side:

-8y = -x-9

We'll divide by -8:

y = x/8 + 9/8

The slope of the line has to be equal to the slope of y = x/8 + 9/8.

m=m2

m = 1/8

The line that has to be determined is:

y = x/8 + n

Now, we know that the f'(x) and the parabola have a common point.

We'll calculate f'(x) = 0

12x-8 = 0

3x-2 = 0

x = 2/3

We'll substitute 2/3 into the equation of the parabola:

y = 24/9 - 16/3 + 9

y = (24+81-48)/9

y = (105-48)/9

y = 57/9

y = 19/3

We'll substitute x and y in y = x/8 + n:

19/3 = 2/24 + n

n = 19/3 - 2/24

n = (19*8-2)/24

n = (19*4-1)/12

n = 75/12

The equation of the line is:

y = x/8 + 75/12

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

The given parabola and the lines are:

y = 6x^2 -8x +9 = 0 and  x-8y = 9.

To find the line which is normal to the parabola and perpendicular to the given line.

The equation of the line perpendicular to the line x-8y = 9 is of the form (interchanging  the coefficents of x and y and putting a minus sign to one of the two) -(-8)x+y = k . Or 8x+y = k.  Or 

 y =-8x+k .........................(1) (slope intercept form) where -8 is the slope and k  is a constant.

Now the slope of the norma to parabola y^2 = 6x^2-8x+9 is  given by the value of 1/dy/dx  at  some  point  is to be infinite . Or dy/dx = 0.

y^2=6x^2-8x+9

2ydy/dx = 12x-8 

dy/dx = y(12x-8) = 0

dy/dx = (12x-8)sqrt(6x^2-8x+9) = 0.

dy/dx = (12x-8)= 0 or 6x^2-8x+9 = 0.

Therefore 12x-8 = 0 and the 6x^2-8x+9 = 0

6x-8 = 0 gives x = 2/3 , to get y , put x=2/3 in y = 6x^2-8x+9:

So y = 6(2/3)^2-8(2/3)+9 = 8/3-16/3+9 = 19/3.

Put x = 2/3  and y = 19/3  i n the lline  y = -8x+k at (1):

19/3 = -8(2/3)+k . Or

k = 19/3 +16/3 = 35/3.

Therefore the required line is: y = -8x+35/3, whic iersects the parabola at (2/3 , 19/3) and is normal to parabola and the given line.

 

 Or 6x^2-8x+9 = 0 has no real solution as the discriminant (-8)^2 - 4*6*9 = -152 which is negative.

So 12x=8

 

 

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