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# Equation of the normal to a curve watch

1. Find the equation of the normal to the curve at the point with the iven x coordinate.

y= 1-2x^2 where x = -2 so i found out y = -7 so the coordiante are (-2,-7)

I then diffrentiated to get -4x and found the gradient of the tanagnt whoch was 8. So the gradient of the normal will be - 1/8

using Y=Mx+C i got y=-1/8 (-2) +C so c is equal to -27/4

Y= 1/4x - 27/4

So then i multiplied by 4 to get rid of the fractions giving

4y=x-27 as the equation, but in the back of the book it says the answer is 8y-x-58 anyone see where i went worng? Thanks.
2. Never mind i relasied after using the Y-y1=M(x-x1) formual how to do it .
3. (Original post by hazbaz)
Find the equation of the normal to the curve at the point with the iven x coordinate.

y= 1-2x^2 where x = -2 so i found out y = -7 so the coordiante are (-2,-7)

I then diffrentiated to get -4x and found the gradient of the tanagnt whoch was 8. So the gradient of the normal will be - 1/8

using Y=Mx+C i got y=-1/8 (-2) +C so c is equal to -27/4

Y= 1/4x - 27/4

So then i multiplied by 4 to get rid of the fractions giving

4y=x-27 as the equation, but in the back of the book it says the answer is 8y-x-58 anyone see where i went worng? Thanks.
You went wrong here:
using Y=Mx+C i got y=-1/8 (-2) +C so c is equal to -27/4
Substituting the (-2.-7) point into the equation of the line:
-7=-1/8(-2)+C-> -29/4

and here:
Y= 1/4x - 27/4
IN the equation of the normal line for tangent use the value of -1/8 calculated at the beginning
y=-1/8x-29/4
THe answer of the book is not correct in that form you wrote it,
or there may be difference in the initial eqution of the curve or any data between you wrote here and given by the book.

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