Question

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?

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

oh! wait hold on a sec lemme draw something

Answer 7

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

Answer 8

oh thanks for visual i like visuals

Answer 9

lol :)

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

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?

Answer 15

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

Answer 16

ok i think this is right lol

Answer 17

are you looking at what i did rain?

Answer 18

yep, trying to figure it out

Answer 19

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

Answer 20

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

Answer 21

from x=0 to x=2cp

Answer 22

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

Answer 23

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

Answer 24

you got 2c^2?

Answer 25

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

Answer 26

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

Answer 27

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

Answer 28

ok cool rain :)

Answer 29

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

Answer 30

good luck Rain, hope you get the correct answer