# Equ. of normal

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• Apr 17th 2006, 11:16 AM
dadon
Equ. of normal
Could you check this question to see if I have done it right:

Question: Find the equation of the normal to the curve y=x+ 4/x at the point p where x=4. If the normal meets the curve agian at N, find the co-ordinates of N.

This is what I have done:
y= x+4x^-1
dy/dx= 1 - 4x^-2
x=4 y=(4)*4(4)^-1 = 5
dy/dx= 1 - 4(4)^-2 = 0.75 or 3/4
Normal gradient = -4/3
(y - 5) = -4/3(x - 4)
y - 5 = -4/3x + 16/3
y = -4/3x + 31/3 or 3y = - 4x + 31

This is where I got up to, i'm not to sure what you have to do when it says 'If the normal meets the curve agian at N, find the co-ordinates of N'

If you could please explain and check if I have done the first bit right.

Thank you
• Apr 17th 2006, 11:29 AM
TD!
Looks good so far, at first sight.

To find all intersection points, solve the equation you get by expressing that the function values have to be equal, i.e. solve:

$\displaystyle x + \frac{4} {x} = - \frac{4} {3}x + \frac{{31}} {3}$

Of course, one of the solutions is x = 4, but you are asked to find possible others (there is one more).
• Apr 17th 2006, 11:30 AM
ThePerfectHacker
Quote:

Originally Posted by dadon
Could you check this question to see if I have done it right:

Question: Find the equation of the normal to the curve y=x+ 4/x at the point p where x=4. If the normal meets the curve agian at N, find the co-ordinates of N.

This is what I have done:
y= x+4x^-1
dy/dx= 1 - 4x^-2
x=4 y=(4)*4(4)^-1 = 5
dy/dx= 1 - 4(4)^-2 = 0.75 or 3/4
Normal gradient = -4/3
(y - 5) = -4/3(x - 4)
y - 5 = -4/3x + 16/3
y = -4/3x + 31/3 or 3y = - 4x + 31

This is where I got up to, i'm not to sure what you have to do when it says 'If the normal meets the curve agian at N, find the co-ordinates of N'

If you could please explain and check if I have done the first bit right.

Thank you

If $\displaystyle y=x+\frac{4}{x}$
Then, $\displaystyle y'=1-4x^{-2}$
At $\displaystyle x=4$ you have,
$\displaystyle y'=3/4$
But you need normal which is perpendicular to tangent thus its slope is,
$\displaystyle m=-4/3$ and passes through point $\displaystyle (4,5)$ (because at $\displaystyle x=4$ we have $\displaystyle y=5$).
Thus, using slope-intercept formula you have,
$\displaystyle y-5=-\frac{4}{3}(x-4)$
Thus,
$\displaystyle y=-\frac{4}{3}x+\frac{31}{3}$
Look like you are right.
• Apr 17th 2006, 12:37 PM
dadon
Quote:

To find all intersection points, solve the equation you get by expressing that the function values have to be equal, i.e. solve:

$\displaystyle x + \frac{4}{x} = - \frac{4}{3}x + \frac{{31}}{3}$

Of course, one of the solutions is x = 4, but you are asked to find possible others (there is one more).

Hey! thanks for replying and checking my answer

so to solve the equ.

3x + 12/x =-4x + 31

3x + 12 = x(-4x + 31)

-4x^2 -28x - 12 = 0

4x^2 + 28x + 12 = 0

x^2 + 7x + 3 = 0

x = -0.45 x = -6.54

Is this right? as you said that one is meant to be x = 4
• Apr 17th 2006, 02:38 PM
TD!
Quote:

Originally Posted by dadon
3x + 12/x =-4x + 31

3x + 12 = x(-4x + 31)

You miss a square in the first term, it should say 3x². Further one you also made a sign mistake.
• Apr 18th 2006, 02:17 AM
dadon
Re:
Quote:

You miss a square in the first term, it should say 3x². Further one you also made a sign mistake.

Hey! So the answer should be:

$\displaystyle 3x + 12/x = -4x + 31 3x^2 + 12 = x(-4x + 31) 3x^2 + 4x^2 - 31x + 12 = 0 7x^2 - 31x + 12 = 0 x = 4, x = 3/7$
• Apr 18th 2006, 04:49 AM
ThePerfectHacker
Quote:

Originally Posted by dadon

Hey! So the answer should be:

$\displaystyle 3x + 12/x = -4x + 31 3x^2 + 12 = x(-4x + 31) 3x^2 + 4x^2 - 31x + 12 = 0 7x^2 - 31x + 12 = 0 x = 4, x = 3/7$

Yes
• Apr 18th 2006, 05:29 AM
dadon
Re:
Quote:

If the normal meets the curve agian at N, find the co-ordinates of N.

So when x = 3/7

y = (3/7) + 4(3/7)^-1

y = 205/21

(3/7, 205/21)

Thanks guys!
• Apr 18th 2006, 09:49 AM
TD!
You're welcome :)