# question4 (1 Viewer)

#### Gruma

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

N

Nope, ab.

#### Gruma

##### Member
smart arse

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

#### walla

##### Satisfied Customer
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

N

#### ND

##### Guest
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

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

#### fitz33

##### Member
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

##### werd!
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!

#### xiao1985

##### Active Member
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

##### N1NJ4
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

##### Member
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

##### ReSpEcTeD
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

##### New Member
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

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

#### MyLuv

##### Member
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

##### Member
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