hon1hon2hon3 said: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 . . .
thewog2004 said: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
Grrrr.cccclaire said: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.