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A Curve has gradient (ax^2+b)/x^2. The equations of the tangents to the curve at x=-1 and x=1 are y=-2 and y=2 respectively. Find the values of a and b and the equation of the curve.

My Solution:

Part 1:

-Lets the equation of tangent be y=mx+c where m is the gradient and c is the intercept and (x,y) is the point

-Given when x=-1, y=-2 from equation of tangent :
-> -m+c = -2 --------- ( 1 )

- Given when x=1, y=2 from equation of tangent :
-> m+c = 2 ------------( 2 )

Solving (1) and (2) simultaneously:
-> m=2 and c=0

Part 2:

-Given dy/dx = (ax^2+b)/x^2=a + bx^-2
-making dy subject of formulae and integrating both sides
-> y= ax-bx^-1+c (Giving equation of curve)

From Part 1, c=0

-> Therefore y=ax-bx^-1 --------------(3)

Part 3:

Given when x = -1 , y=0 from curve
-> therefore a = - b -------------------(4)

Given when x=-1, y= -2 on tangent:
-> therefore -a + b = -2 --------------(5)

Solving (4) and (5)

-> a= 1 and b=-1 ( which is by the way the answer in the book )

My question is: IS the solution right??? because the equation ( 4 ) i got it by inventing which is what i doubt cannot be good even though the answers are right. IF my reasoning is wrong can someone help me do the problem in there own way!!!

Last edited by tyre; 06-05-2012 at 20:12.
2. Re: Doubt about my solution
Someone help me clear this doubt please!!!!!!!!!!
3. Re: Doubt about my solution
Is my solution correct or not?? If yes please explain !!!
4. Re: Doubt about my solution
5. Re: Doubt about my solution
Hi, I think you've gotten lucky with the answers there! The c in the y = mx + c equation doesn't equal the constant of integration.

You don't need step 1 there at all, you're already given the equations of the tangents! Your tangents are
y = -2
y = 2
These are horizontal lines, meaning their slope is 0. Think turning points. What can you say about the derivative of a function at a turning point? You can get an equation in a and b here.

You're correct to integrate the gradient of the curve.
Your equation of the curve is correct.
So you can sub in your values of x and y to get two equations in a, b and c here. Then eliminate c, to get another equation in a and b.
Last edited by finality; 06-05-2012 at 21:52.
6. Re: Doubt about my solution
just to point out that if you have two tangents you have two different m values and two different c values, so you can't use simultaneous equations to find them
7. Re: Doubt about my solution
Thnks to Finality and the bear for helping!!! I will try what yu guys suggest then will post back my working ok???
8. Re: Doubt about my solution
Here is my working:

So since the tangent are y=-2 and y= 2 i.e horizontal lines
->dy/dx=0
->therefore (ax^2+b)/x^2=0
-> a + bx^-2=0 ------------------( 1 )

When x = -1 or x= 1, from (1)
-> a = - b -------------------------( 2 )

Since equation of curve is: ax-bx^-1+c --------------( 3 )
From ( 3 ), when x=-1, y=-2

-> -a + b + c = -2 ---------------------( 4 )

From ( 3 ) , when x=1, y=2

-> a - b + c = 2 -----------------------( 5 )

Solving ( 4 ) and ( 5 ) simultaneously

-> 2b-2a=-4
-> b-a = -2 ---------------------( 6 )

Solving ( 6 ) and ( 2 )

-> a = 1 and b = -1 ( which is the answers)

So thnks yu again both of yu for helping me

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Last updated: May 6, 2012
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