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

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- anonymous

Asymptotes 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

this sounds like you may need integration? do you know integration?

- anonymous

Yep. I just don't know how to apply it to this question

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- anonymous

as 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

well 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

oh! wait hold on a sec lemme draw something

- anonymous

##### 1 Attachment

- anonymous

am 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

oh thanks for visual i like visuals

- anonymous

lol :)

- myininaya

you need to find the x-intercept of the tangent line

- anonymous

alright, I think I know how to do this now. Thanks!

- myininaya

i will do it too and we can compare answers k :)

- anonymous

okay :) i'll come back when I'm done

- anonymous

oh... 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

i didnt check this but you said to find tangent at (cp,c/p) right ?

##### 1 Attachment

- myininaya

ok i think this is right lol

- myininaya

are you looking at what i did rain?

- anonymous

yep, trying to figure it out

- anonymous

I integrated the wrong thing. I put the equation of the hyperbola instead

- myininaya

right according to your pic we are just wanting the area under that tangent line

- myininaya

from x=0 to x=2cp

- anonymous

y=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

ah, okay, I got the answer :) Awesome, thanks so much!!

- myininaya

you got 2c^2?

- anonymous

I haven't finished yet, on the third last line. it's just algebra, so it should be right

- anonymous

okay, got the answer :D
Thank you!!!

- myininaya

on that attachemnt i started my work in the second colum n and finished upp inthe first

- myininaya

ok cool rain :)

- anonymous

since m=slope=-c^2/x^2 you can get the EQ of the line
y-y1=m(x-x1)
y- c/p = (-c^2/ x^2)(x-cp)
y = -c^2/x +c^3 p/x^2 +c/p you can now do the area by integration here

- anonymous

good luck Rain, hope you get the correct answer

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