Differentiation and Tagents/Normal help!?

Question

Find the equation of the NORMAL to the curve y= 4x^2 - 7x + 2 at the point (2,4)

Answer 1

Differentiate the equation of the curve y= 4x^2 - 7x + 2

dy/dx = 8x - 7

-dx/dy = 1/(7-8x)

slope of the normal is -dx/dy =1/(7-8x)
(This can be seen from the fact that slope of tangent * slope of normal = (dy/dx)*(-dx/dy) = -1......which is true for any pair of perpendicular lines!)

Let the normal be y=mx + c

m = 1/(7-8*2) = -1/9

It passes through the point (2,4): thus the point satifies the equation of the normal.

4 = -(1/9)*2 + c
c = 38/9

Thus normal at (2,4) is y = -(1/9)*x + (38/9)

x + 9y = 38 is the required normal!

Answer 2

The slope of the tangent to the curve is y', so:

y' = 8x - 7.

At point (2,4), you have y'(2) = 9, which is the slope of the tangent line. The normal is perpendicular on the tangent, so its slope is -1/9 (the product of the slopes of two perpendicular lines is -1). The equation of the normal will then be of the form:

y = -1/9x + n.

You have to find n now. You know that the point (2,4) is on the normal line, so:

4 = -2/9 + n ==> n = 38/9

Therefore, the equation of the normal will be:

y = -1/9x + 38/9.

Answer 3

dy/dx = 8x - 7
dy/dx = 16 - 7 = 9 at point (2,4)
gradient of perpendicular at (2,4) = -1/9
perpendicular passes through (2,4) Equation of perpendicular is given by:-
y - 4 = (-1/9)(x - 2)
y = (-1/9)x + 2/9 +4
y = (-1/9)x +2/9 + 36/9
y = (-1/9)x + 38/9

Answer 4

the equation of the normal to
the curve is
y-y1={-1/(dy/dx)1(x-x1)
where (dy/dx)1is dy/dx at x1

let point(x1,y1)=(2,4)

dy/dx=8x-7
(dy/dx)1=8*2-7=9

therefore,the equation of the
normal to the curve
y= 4x^2 - 7x + 2 at the point (2,4)
is y-4= -1/9(x-2)
>>>>>9y+x-38=0

i hope that this helps