1. ## 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

2. 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).

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.

4. 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

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.

6. ## Re:

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
$

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

8. ## Re:

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!

9. You're welcome