question4 (1 Viewer)

[Gruma]

did ne one else prove q4 (b) [111] to be ab/2 instead of ab?

 

[ND]

Nope, ab.

 

[Gruma]

smart arse

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

 

[walla]

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

 

[ND]

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.

 

[Constip8edSkunk]

i used 1/2 OA*OB*sin BOA... prolly not the most elegant way, but got ab

 

[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

 

[Toodulu]

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]

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]

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!

 

[fitz33]

Originally posted by freaking_out
what? i never heard of a formula for that!

this website has the formula:

http://mathworld.wolfram.com/TriangleArea.html

came in handy today!

 

[xiao1985]

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]

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]

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

 

[Dash]

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

 

[hagun]

Originally posted by nerdd
yah and 90 was the @...........

therfore 1/2bh

How do u know that @ is 90?
The angle formed is 2(b tan@/a sec@) rite?

 

[fitz33]

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

 

[MyLuv]

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]

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