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=?

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