IIT JEE , AIEEE Forum - Mathematics , Analytical Geometry

ashish_banga

if two of the lines $a{x}^{3} + b {x}^{2}y + cx{y}^{2} + d{y}^{3}$ = 0 makes complementary angles wuth the x axis in the anticlockwise sense, then find the relation between a , b ,c ,d

ans= a( a - c ) = d ( d -b )

ashish_banga

aatish please solve u can
and all others also

Aatish

I m trying buddy....but m not sure that i m getting it.........

computer001

simple enuf:

the given eqn represents 3 lines passing thru origin

$y={m}_{1}x,y={m}_{2}x,y={m}_{3}x$

devide eqn by ${x}^{3}$

rewrit as:

$a+bm+c{m}^{2}+d{m}^{3}$

where roots are ${m}_{1},{m}_{2},{m}_{3}$

$\sum{m}_{i}=-c/d$

$\sum{m}_{i}{m}_{j}=b/d$

${m}_{1}{m}_{2}{m}_{3}=-a/d$

u no relation between ${m}_{1},{m}_{2}$ ...(given condition)

solve and eliminate ${m}_{3}$ also...

ashish_banga

can u please tell the relation between m1 and m2

computer001

${m}_{1}{m}_{2}=1$

u get tht using ur argand plane diagram

ashish_banga

please explain fully yaar
NOT UNDERSTOOD

computer001

Re:what a question

most.happening

(y-m1x)(y-m2x)(y-m3x)= the given eq.

so we can say that m1.m2.m3=-a/d

and summ. m1m2=b/d

and m1+m2+m3= -c/d

since m1 m2 are complementary we can say that m1.m2=1

so frm the above three relations we easily get the required result.........

its not so difficult ...........may be u r a begginer FOR THIS TOPIC

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