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Finding equation of a normal - solved

I've tried doing this question but I keep getting it wrong: Answer is :
12y= 12x - 17

- Find the equation of the normal to the curve y=3x^2 - 2x - 1 which is parallel to the line y=x-3 .

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I start off by finding dy/dx , which is 6x-2 . I set it equal to 1 (gradient of the parallel) to find the x co-ordinate.

6x -2 = 1
6x=3
x= 0.5
Y= -5/4

Plug 0.5 back in to find the gradient, it gives 0 :confused:

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(edited 13 years ago)

Reply 1

chembob

Remember you're trying to find the normal, not the tangent.

The gradient of the normal is perpendicular to that of the tangent.

Reply 2

Gelato

Original post by chembob

Remember you're trying to find the normal, not the tangent.

The gradient of the normal is perpendicular to that of the tangent.

Don't I do that after I get the gradient of the tangent at x ? then reciprocate it ?
So it'd still be 0 ?

Reply 3

$Avatar for Get me off the £\?%!^@ computer$

Get me off the £\?%!^@ computer

You're not finding the gradient. You're finding out where it equals 1. This is when x=1/2 and y=-5/4.

So at (1/2,-5/4) the curve has a tangent with a gradient of 1. The normal here has a gradient of -1.

Reply 4

Gelato

Original post by Get me off the £\?%!^@ computer

Okay so (1/2,-5/4) are the points, -1 is the gradient.

Putting it in the formula of straight line :

y+5/4 = -1 (x - 1/2)

Multiply through by 4 to give:
4y+5 = -4x - 2

4x+4y+7 =0

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Answer is meant to be 12y= 12x - 17 :confused:

Reply 5

$Avatar for Get me off the £\?%!^@ computer$

Get me off the £\?%!^@ computer

Original post by Gelato

Okay so (1/2,-5/4) are the points, -1 is the gradient.

Putting it in the formula of straight line :

y+5/4 = -1 (x - 1/2)

Multiply through by 4 to give:
4y+5 = -4x - 2 <-----------------------------------should be plus

4x+4y+7 =0

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Answer is meant to be 12y= 12x - 17 <----------------------------------- this clearly has the wrong gradient.

:confused:

Both answers are wrong but yours is better.

Reply 6

Gelato

Argh not sure :frown:

That's the answer in the back of the book

(edited 13 years ago)

Reply 7

gdunne42

Original post by Gelato

Argh not sure :frown:

That's the answer in the back of the book

Your coordinates are wrong.

The gradient of the normal = 1 so the gradient of the tangent (perpendicular) at the required point = -1
giving

dy/dx = 6x - 2 = -1

so x= 1/6 and you can substitute into the original equation to find y

(edited 13 years ago)

Reply 8

Gelato

Thanks for the help , I see where I went wrong .

I got x as 1/6 Y as -7/6 gradient as -1

Putting them in the formula gives:

y+7/6 = -1 ( x - 1/6 )

Multiplying through by 6 on the left:
6y + 7 = -6 (x - 1)
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I end up with 6x+ 6y = 13

(edited 13 years ago)