I've tried to solve this problem for over 3 or 4 hours but I can't figure out how, please help me.

Ok so it's a curve ( parable? ) y = (ax - b) / ( 1 - x^2 )

You are supposed to solve what 'a' and 'b' should be if the curve goes through the point (-2,1) AND that the tangents* direction of that same point (-2,1) is 45 degrees ( so the tangents 'k' value should be 1, wich is 45 degrees ).

*I'm not sure but I might have the wrong name in English for 'tangent', but it is a line with the same 'k' value as the single point in the curve... if you know what I mean?

THE BELOW IS WHAT I'VE TRIED TO DO:

Ok, so we all know that to get the 'k' value of a point in a curve you must begin with taking the derivate of the curve, wich in this case would be:

MAOL s.43 : D f/g = ( gDf - fDg ) / ( g^2 )

#1. y' = ( ( 1-x^2 )*a - ( ax-b )*( 2x ) ) / ( 1 - x^4 )

and then you should put the derivate in a function of the x value, something like: y'(x) = ... , wich would in this case be: y'(-2) = ... .

so we take that y'(-2) = ... and equal it to 1, because that's the 'k' value we had to have for that point.

and thus we should get an equation with both 2 unknown variables, 'a' and 'b' , in wich we should somehow figure out what they should be to get this to work... wich is where I fail.

The answer for this problem is that 'a' and 'b' both = 1.

I've tried to do the countings myself etc but I can't come up with it...

EDIT:

And could someone clarify why I got an Infraction for spamming Advertisements in this post?? As far as I know I havn't adversided anything!
????

2. Originally Posted by me_maths
I've tried to solve this problem for over 3 or 4 hours but I can't figure out how, please help me.

Ok so it's a curve ( parable? ) y = (ax - b) / ( 1 - x^2 )

You are supposed to solve what 'a' and 'b' should be if the curve goes through the point (-2,1) AND that the tangents* direction of that same point (-2,1) is 45 degrees ( so the tangents 'k' value should be 1, wich is 45 degrees ).

*I'm not sure but I might have the wrong name in English for 'tangent', but it is a line with the same 'k' value as the single point in the curve... if you know what I mean?

THE BELOW IS WHAT I'VE TRIED TO DO:

Ok, so we all know that to get the 'k' value of a point in a curve you must begin with taking the derivate of the curve, wich in this case would be:

MAOL s.43 : D f/g = ( gDf - fDg ) / ( g^2 )

#1. y' = ( ( 1-x^2 )*a - ( ax-b )*( 2x ) ) / ( 1 - x^4 )

and then you should put the derivate in a function of the x value, something like: y'(x) = ... , wich would in this case be: y'(-2) = ... .

so we take that y'(-2) = ... and equal it to 1, because that's the 'k' value we had to have for that point.

and thus we should get an equation with both 2 unknown variables, 'a' and 'b' , in wich we should somehow figure out what they should be to get this to work... wich is where I fail.

The answer for this problem is that 'a' and 'b' both = 1.

I've tried to do the countings myself etc but I can't come up with it...

Your method is good, but in your derivative you've got a sign problem and you squared the denominator incorrectly.
$\displaystyle y = \frac{ax - b}{1 - x^2}$

$\displaystyle y' = \frac{(a)(1 - x^2) - (ax - b)(-2x)}{(1 - x^2)^2}$

Note the "-" on the (-2x) factor. Also $\displaystyle (1 - x^2)^2 = 1 - 2x^2 + x^4$, NOT $\displaystyle 1 - x^4$.

-Dan

3. Originally Posted by topsquark
Your method is good, but in your derivative you've got a sign problem and you squared the denominator incorrectly.
$\displaystyle y = \frac{ax - b}{1 - x^2}$

$\displaystyle y' = \frac{(a)(1 - x^2) - (ax - b)(-2x)}{(1 - x^2)^2}$

Note the "-" on the (-2x) factor. Also $\displaystyle (1 - x^2)^2 = 1 - 2x^2 + x^4$, NOT $\displaystyle 1 - x^4$.

-Dan
ahh, thanks for the help topsquark

but it still leaves 2 unknown variables in the equation.

I cleaned up the equations and I got it to be:

$\displaystyle y'(-2) = \frac{a(1-4)-(-2a-b)(4)}{1-8+16}$
<=>
$\displaystyle y'(-2) = \frac{5a+4b}{9}$

so I put that equal to 1, wich was the needed angle ( k value ) of the point on the curve:

$\displaystyle y'(-2) = \frac{5a+4b}{9} = 1$

and here is where i'm stuck then...

I get $\displaystyle a = \frac{9-4b}{5}$ etc, wich doesn't really help at all now does it?

so, where should I continue from here?

Once again, the solution to the problem is that 'a' and 'b' should both equal 1, but I have problems in getting there myself , help is much appreciated!

4. Ahh, now I've got it!

I take the first equation:

$\displaystyle y = \frac{ax - b}{1 - x^2}$

put in the point it should be on (-2,1)

$\displaystyle -2a - b = -3$

wich makes it:

$\displaystyle -2a - b = -3$ or $\displaystyle 2a + b = 3$, wichever you want.

As I said, I take the first equation and compare it to the result I got in the last post:

$\displaystyle 1 = \frac{5a + 4b}{9}$

<=>

$\displaystyle 5a + 4b = 9$

Then I take both of them and compare them:

#1. $\displaystyle 2a + b = 3$
#2. $\displaystyle 5a + 4b = 9$

I'll use the addition method so I take #1. times (-4) and I get:

#1. $\displaystyle -8a - 4b = -12$
#2. $\displaystyle 5a + 4b = 9$

then I kill the b's and I'm left with:

#1. $\displaystyle -8a = -12$
#2. $\displaystyle 5a = 9$

wich then is added to become:

$\displaystyle -3a = -3$
<=>
$\displaystyle -a = -1$
<=>
$\displaystyle a = 1$

and then I just put that a = 1 into for instance here:

#1. $\displaystyle 2a + b = 3$
<=>
#1. $\displaystyle 2 + b = 3$
<=>
#1. $\displaystyle b = 3-2$
<=>
#1. $\displaystyle b = 1$

Voila, thanks for all the help guys