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Gruma

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smart arse

i used 1/2 x base x height
using base as acosx / 1 + sinx
and height as bcosx / 1 - sinx
 

walla

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i used base as OB
height was perpendicular distance from (y=-bx/a - negative asymptote) to A
then 1/2 base times height
worked out to ab
 
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ND

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I broke the area up into the area of teh triangles AOT and ABT where T was the pt of intersection of the tangent and x-axis.
 

fitz33

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the elegant way to do it was to use the formula for the area enclosed between three points..just apply the formula and thats it, only takes a few lines
 

Toodulu

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Originally posted by Gruma
i used 1/2 x base x height
using base as acosx / 1 + sinx
and height as bcosx / 1 - sinx
that's what i did but must have gotten some algebra wrong on the way
 

freaking_out

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Originally posted by fitz33
the elegant way to do it was to use the formula for the area enclosed between three points..just apply the formula and thats it, only takes a few lines
what? i never heard of a formula for that! :(
 

freaking_out

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Originally posted by fitz33
the elegant way to do it was to use the formula for the area enclosed between three points..just apply the formula and thats it, only takes a few lines
what? i never heard of a formula for that! :(
 

xiao1985

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Originally posted by Gruma
smart arse

i used 1/2 x base x height
using base as acosx / 1 + sinx
and height as bcosx / 1 - sinx

this only works if the hyperbola is a rectangular hyperbola, that is the asymptotes intersect at right angles..... i fink it is more apropriate to use the area formula:

a = 1/2 ab sin(@) or something, where a and b are two sides and @ is the angel between the sides
 

underthesun

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This question was once asked by the wise old man, from the mountains of extension 2 math..

ok, this question was once posted in the ext2 forum by oldman. After looking at ND's response back then, i learned how to do it the easy way. So fortunate that i learned something back then...
 

nerdd

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Originally posted by xiao1985
a = 1/2 ab sin(@) or something, where a and b are two sides and @ is the angel between the sides
yah and 90 was the @...........

therfore 1/2bh :D :D
 

Dash

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That last question of 4b was so stupid!!

It agez!! It was so prone to errors!!

I don't even know what I got as the answer cos everything was so messy :(
 

fitz33

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the angle formed isnt 90...its not a rectangular hyperbola
 

MyLuv

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yep,the angle isnt 90.I found tan@ then convert to sin@ then eliminate them,take long time tho,no wonder why i run out of time:(
 

smeyo

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yeah thats proably why i ran out of time to i did the tan@ then converted to sin@ from pathagoras which took a while then the OA and OB distances then x them, luckily it turned out otherwise it would have been the biggest waste of time, it already was a bit, took time away from other questions, ohwell
 

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