Thank you so much. I will try to implement these steps from now.(Original post by duncanjgraham)
Sub in really big postive X values eg X=100  see where the curve approaches
Sub in really negative X values eg X= 100 see where the curve approaches
Sub in an X value extremely close to one side of a known asymptote , eg if an asymptote is X=1 sub in X=1.001 and see where the curve is going
Sub in an X value extremely close to the OTHER side of the known asymptote , eg for the above you'd sub in X=0.009 and this gives you a very very very good idea of the nature of the curve either side of the asymptotes
You then find points of intersections and any stationary points, and this is enough information to then intuitively join up the curve correctly
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 26062016 12:53

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 26062016 12:55
(Original post by tangotangopapa2)
June 2013, Q 8) A line (x + 2y = 2) cuts curve with polar equation (r=1+cos theta ) is two parts. Find the ratio of areas above and below the line enclosed by the curve.
Mark Scheme does not seem to be helpful either. It simply states points of intersection as (0,1) and (2,0) out of nowhere and then proceeds in alien fashion. Could someone explain to me? I don't even know where to begin.
The Cartesian equation of curve is x^{2 + }y^{2 } x = sqrt( x^{2 + }y^{2}).
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 26062016 13:05
(Original post by tangotangopapa2)
Thank you so much. I will try to implement these steps from now.
(You can then put k back into anything important like coordinates AFTER you've got the shape of the graph correct)
This is what I do anyways 
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 26062016 14:31
(Original post by tangotangopapa2)
June 2013, Q 8) A line (x + 2y = 2) cuts curve with polar equation (r=1+cos theta ) is two parts. Find the ratio of areas above and below the line enclosed by the curve.
Mark Scheme does not seem to be helpful either. It simply states points of intersection as (0,1) and (2,0) out of nowhere and then proceeds in alien fashion. Could someone explain to me? I don't even know where to begin.
The Cartesian equation of curve is x^{2 + }y^{2 } x = sqrt( x^{2 + }y^{2}). 
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 26062016 14:41
(Original post by becksreills)
I was very confused on this question yesterday and I'm still not sure how the mark scheme has done it. The way I did it was convert x+2y=2 into polar coordinates and draw it with the other graph since it cuts it at theta= 0 and pi/2 when you draw it like that, then it becomes a lot clearer which areas you need to find! 
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 26062016 14:45
(Original post by tangotangopapa2)
Thank you, but how do you convert x+2y=2 to polar coordinates. (Also this line does not pass through the origin.)
It doesn't pass through the origin, but the value of r at theta=0 is the same as the value of r at theta=0 for the other graph so it cuts it there 
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 26062016 15:16
How do I do 7iii. I know the solution but could somebody explain why the answer is what it is.. I have found the point in polar coordinates of P, but can't see how this helps me find the tangent...

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 26062016 15:18
(Original post by becksreills)
x=rcos(theta) and y=rsin(theta) so sub those in and rearrange to r=f(theta)
It doesn't pass through the origin, but the value of r at theta=0 is the same as the value of r at theta=0 for the other graph so it cuts it there
So, by substituting x=rcos(theta) and y=rsin(theta) and rearranging to r=f(theta), I got r = 2/(cos(theta) + 2 sin (theta)). Then equating 'r' with 'r' of r=cos(theta)sin2(theta). This equation is very hard to solve. (I cant ) but as you said, theta = 0 and pi/2 seem to satisfy the equation. Now how do we find the required areas.
Clearly, finding total area enclosed by curve and area between theta = 0 and pi/2 won't give the required areas, as area of sector (i.e enclosed by curve and half lines theta = 0 and pi/2) is not what we are interested in. We are interested in areas cut by lines. I couldn't find a way to proceed. 
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 26062016 15:23
(Original post by Mathematicus65)
How do I do 7iii. I know the solution but could somebody explain why the answer is what it is.. I have found the point in polar coordinates of P, but can't see how this helps me find the tangent... 
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 26062016 15:28
IN JUNE 2012 YOU ARE MADE TO FACTORISE A CUBIC EQUATION IN TWO DIFFERENT QUESTIONS  wtf? I don't even know how to do that **** methodically?
*edit* it's a paper with low grade boundaries and you don't lose too many marks for not being able to factorise the cubic , still annoying tho: PLast edited by duncanjgraham; 26062016 at 15:31. 
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 26062016 15:28
(Original post by tangotangopapa2)
The line of symmetry is y=x. At P, (intuitively) the gradient of the tangent should be perpendicular to the line of symmetry. So the gradient at this point is 1. Now the equation of line passing through P and having gradient 1 is the required equation of line. 
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 26062016 15:29
(Original post by tangotangopapa2)
Thank you
So, by substituting x=rcos(theta) and y=rsin(theta) and rearranging to r=f(theta), I got r = 2/(cos(theta) + 2 sin (theta)). Then equating 'r' with 'r' of r=cos(theta)sin2(theta). This equation is very hard to solve. (I cant ) but as you said, theta = 0 and pi/2 seem to satisfy the equation. Now how do we find the required areas.
Clearly, finding total area enclosed by curve and area between theta = 0 and pi/2 won't give the required areas, as area of sector (i.e enclosed by curve and half lines theta = 0 and pi/2) is not what we are interested in. We are interested in areas cut by lines. I couldn't find a way to proceed.
Here's my diagram:
And here's my working if you're interestedSpoiler:ShowLast edited by Parallex; 26062016 at 15:34.Post rating:1 
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 26062016 15:36
(Original post by duncanjgraham)
IN JUNE 2012 YOU ARE MADE TO FACTORISE A CUBIC EQUATION IN TWO DIFFERENT QUESTIONS  wtf? I don't even know how to do that **** methodically?
*edit* it's a paper with low grade boundaries and you don't lose too many marks for not being able to factorise the cubic , still annoying tho: P 
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 26062016 15:38
(Original post by Parallex)
You don't need to convert x+2y=2 in to polar form. Draw the polar function onto an xy grid and you can see where it intersects.
Here's my diagram:
And here's my working if you're interestedSpoiler:Show 
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 26062016 15:38
(Original post by duncanjgraham)
IN JUNE 2012 YOU ARE MADE TO FACTORISE A CUBIC EQUATION IN TWO DIFFERENT QUESTIONS  wtf? I don't even know how to do that **** methodically?
*edit* it's a paper with low grade boundaries and you don't lose too many marks for not being able to factorise the cubic , still annoying tho: P
Posted from TSR Mobile 
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 26062016 15:39
(Original post by marioman)
I don't remember these questions? Which ones are you talking about? 
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 26062016 15:40

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 26062016 15:40
(Original post by tangotangopapa2)
Thank you so, so much. Brilliant solution. 
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 26062016 15:44
(Original post by duncanjgraham)
june 2012 3ii and 8iii
Posted from TSR Mobile 
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 26062016 15:45
(Original post by duncanjgraham)
june 2012 3ii and 8iii
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Updated: June 29, 2016
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