Solve Tricky Problem for Curve y=(ax-b)/(1-x^2)-SOLVED by courtrigrad"

[me_maths]

tricky problem please HELP :( -SOLVED by "courtrigrad"-

[SOLVED, thanks to "courtrigrad"!]

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, which 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, which 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) = ... , which 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 which we should somehow figure out what they should be to get this to work... which 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...

Could someone please help me!?

 

[courtrigrad]

You know that $$1 = \frac{-2a-b}{-3} \Rightarrow -2a-b = -3$$Also $$y'(x) = \frac{(1-x^{2})(a)-(ax-b)(-2x)}{(1-x^{2})^{2}}$$Thus $$y'(-2) = 1 = \frac{-3a+8a+4b}{9} \Rightarrow 5a + 4b = 9$$

So the two equations are: $$-2a - b = -3$$ and $$5a+4b = 9$$

Solve these two equations for $$a,b$$ and you should get the answer.

 

[me_maths]

Aah!

Thanks a lot!

Forgot completaly that you could use the first equation to compare it to the result of y'(-2) = ... . Not to mention that I made a few mistakes in the calculating also such as (1-x^2) would be (1-x^4) which is totally wrong...

Thanks again, helped me tremendously! <3 <3