Equ. of normal

• April 17th 2006, 11:16 AM
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
(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
• April 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:

$
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).
• April 17th 2006, 11:30 AM
ThePerfectHacker
Quote:

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

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

$
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).

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
• April 17th 2006, 02:38 PM
TD!
Quote:

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.
• April 18th 2006, 02:17 AM
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:

$
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
$
• April 18th 2006, 04:49 AM
ThePerfectHacker
Quote:

Hey! So the answer should be:

$
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
• April 18th 2006, 05:29 AM
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!
• April 18th 2006, 09:49 AM
TD!
You're welcome :)