Q3-4 hard? (2 Viewers)

[lordesf]

Hey

man wats with it this yr.... Q3-4 were so hard... those circle ones? did anyone get it out?

i really lost all direction when i saw it so i skiped it after like 5 mins looking at it.

 

[Js^-1]

Really? I'm pretty sure I got 100% on those questions. It was question 4 (a) if anyone is interested.

 

[hon1hon2hon3]

In Question 3 , yeah i got a few that i would lose marks . . but question 4 is alright . . hopefully.

For part (a) part (i) = kr /2

part (ii) similar to part (i) seperate the big triangle in to triangle OLM , LOK and KOM and add them together . . soo it becomes triangle KLM = kr/2 + mr/2 + lr/2 = Pr/2.

As for the part (iii) i forgot what did i do = = but i got that its 6.

And for (iv) i got that secound circle is also 2. . . althought it dosent seem right lol . . .

 

[elliots]

In Question 3 , yeah i got a few that i would lose marks . . but question 4 is alright . . hopefully.

For part (a) part (i) = kr /2

part (ii) similar to part (i) seperate the big triangle in to triangle OLM , LOK and KOM and add them together . . soo it becomes triangle KLM = kr/2 + mr/2 + lr/2 = Pr/2.

As for the part (iii) i forgot what did i do = = but i got that its 6.

And for (iv) i got that secound circle is also 2. . . althought it dosent seem right lol . . .

i also got 6 and 2

its about equating area's... 1/2(Pr) = 1/2(base*height)

 

[Js^-1]

Yeah, I got 2 for the second circle too. I love the NOT TO SCALE addition.

 

[yorkstanham]

yeah i also got 6 and 2. yew!

 

[jet]

I got 6 and two as well

 

[midifile]

Yep thats why I got. =]

They werent that hard when you worked out what you had to do

 

[cccclaire]

Yeh the circle triangles looked tricky. In reading time I was like "wtf".

Once you actually did it though you realised it was piss easy.

 

[thewog2004]

did anyone get that OTM Collinear question?

 

[Js^-1]

Yeh, I just the got the equation T satisfied, then showed that (0,0) also satisfied it. Therefore, since the equation was linear, it must be OT. The find the midpoint, and sub it in to show it satisfies that equation.

 

[leoyh]

yeah i did. prove that M and O line on the line that T lines on, which is given in ii

 

[cccclaire]

did anyone get that OTM Collinear question?

Yeh. You just proved that the points for (0,0) and the midpoint satisfied the equation of part (iii), which they did.

 

[thewog2004]

how did u show that the midpoint satisfied the equation...i got sumfin weird like X^2 + Y^2/2a^2 etc :S:S:S:S:S:S

 

[leoyh]

how did everyone get 2 for the radius question. i keep seeming to get 3 o__o

A = 1/2Pr
1/2 (8x15) = 1/2 (8+15+17)r
therefore, r = 3?

 

[cccclaire]

how did u show that the midpoint satisfied the equation...i got sumfin weird like X^2 + Y^2/2a^2 etc :S:S:S:S:S:S

You subed it in and got difference of two squares, then you expanded the top of the fraction and put the ones with x1's and y1's on one side and x2's and y2's on the other side, which satisfied the original equation of the hyperbola (seeing as x1,y1 and x2,y2 lie on the hyperbola.)

Well thats what I did.

 

[hon1hon2hon3]

That midpoint one , i got like a lot of things , but when u seperate them and collect like terms , u get like something xx1/a square + yy1/b square = xx2/a square + yy2/b square . . . which both equal to 1, soo it satisfy the equation.

Dont know if its right, thats just how i did it.

 

[midifile]

The area was 36, not 60

 

[Js^-1]

Yeh...then you have to say that x1 and y1 satisdy the equation of the hyperbola, therefore 1/2(1) -1/2(1) = 0
Which is true.

 

[Poad]

You subed it in and got difference of two squares, then you expanded the top of the fraction and put the ones with x1's and y1's on one side and x2's and y2's on the other side, which satisfied the original equation of the hyperbola (seeing as x1,y1 and x2,y2 lie on the hyperbola.)

Well thats what I did.

Grrrr.

Didnt get that bit. :\