IIT JEE , AIEEE Forum - Mathematics , Algebra - just 2 mins.........help me out!!!!!!!!!!

shakirshafi12

sorry for not typing the solution(it was too late night)

just

putting sol in some time.

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a(a^2 - 3b^2) = 2 ;
b(3a^2 - b^2) = 11;

now as i had said put a=rsinQ

b=rcosQ

we get

r³SinQ(1-4cos²Q)=2-------1

r³CosQ(3-4cos²Q)=11----------2

divide eq 1 and 2

tanQ(1-4cos²Q)=2/11(3-4cos²Q)

now gather terms of cos²Q

we get

(tanQ-6/11)=4cos²Q(tanQ-2/11)

sec²Q(tanQ-6/11)=4(tanQ-2/11)

put sec²Q=1+tan²Q

now put tanQ=t

we get the equation

11t³-6t²-33t+2=0

now put it like this

11t³-22t²+16t²-32t-t+2=0

11t²(t-2)+16t(t-2)-(t-2)=0

so t=2 is a factor

that means tanQ=2

sinQ=2/sqrt(5) cosQ=1/sqrt(5)

using this we conclude that answer is 5