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anonymous
 5 years ago
Prove that the area bounded by a tangent at P(cp, c/p) to the rectangular hyperbola xy = c² and the asymptotes of the curve is a constant (ie, not dependent on p)
Help please?
anonymous
 5 years ago
Prove that the area bounded by a tangent at P(cp, c/p) to the rectangular hyperbola xy = c² and the asymptotes of the curve is a constant (ie, not dependent on p) Help please?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Asymptotes are the x and y axis, and the tangent at P is x + p²y = 8p Other than that, I don't know how to do it

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.1this sounds like you may need integration? do you know integration?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yep. I just don't know how to apply it to this question

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0as x approaches infinity, the hyperbola approaches 0, but it doesn't touch, so I'm not sure what I should integrate since I only have the lower limit

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.1well that tangent line touches the curve sometime after x=0 it touches the tangent at x=cp so assume cp>0 and use as upper limit integrate tanget line  y=c^2/x from 0 to cp and see what happens

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh! wait hold on a sec lemme draw something

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0am I trying to find the yellow bit? I was trying to work out the area of the green bit and... yeah, it didn't work

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.1oh thanks for visual i like visuals

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.1you need to find the xintercept of the tangent line

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0alright, I think I know how to do this now. Thanks!

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.1i will do it too and we can compare answers k :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay :) i'll come back when I'm done

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh... i'm using x + p²y = 8p as the equation of the tangent... that's not right is it... I just realised that I got that from a question where c² = 16 so do I have to work out a general one?

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.1i didnt check this but you said to find tangent at (cp,c/p) right ?

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.1ok i think this is right lol

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.1are you looking at what i did rain?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yep, trying to figure it out

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I integrated the wrong thing. I put the equation of the hyperbola instead

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.1right according to your pic we are just wanting the area under that tangent line

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0y=c^2 /x =c^2 x^(1) slope m= tangent line=c^2 x^(2)= c^2 /x^2 the slope at P(cp, c/p) m=c^2 / c^2p^2=1/p^2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ah, okay, I got the answer :) Awesome, thanks so much!!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I haven't finished yet, on the third last line. it's just algebra, so it should be right

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay, got the answer :D Thank you!!!

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.1on that attachemnt i started my work in the second colum n and finished upp inthe first

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0since m=slope=c^2/x^2 you can get the EQ of the line yy1=m(xx1) y c/p = (c^2/ x^2)(xcp) y = c^2/x +c^3 p/x^2 +c/p you can now do the area by integration here

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0good luck Rain, hope you get the correct answer
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