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Finding equation of a normal - solved Watch

1. 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 .

----------------------

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

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

The gradient of the normal is perpendicular to that of the tangent.
3. (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 ?
4. 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.
5. (Original post by 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.

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
6. (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

----------

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

Both answers are wrong but yours is better.
7. Argh not sure

That's the answer in the back of the book
8. (Original post by Gelato)
Argh not sure

That's the answer in the back of the book

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
9. 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)
-------------------------

I end up with 6x+ 6y = 13

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Updated: December 17, 2010
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