If the line y=3^1/2x cuts the curve X⁴+ax²y+bxy+cx+dy+6=0 at A,B, C,D then OA.OB.OC.OD(where O is the origin) is

 y= root(3)x  is cutting the curve at four points....A,B,C,D 

and O is the origin...

You have to find out OA.OB.OC.OD

Our approach will be some thing like this....

Let OA be r..means distance OA is r...

O is origin

SO we can now find the co-ordinate of A...

it will be  (rcosa , rsina)    ...where tan a is slope 

hence A ( r/2 ,r.root(3)/2 )

Now this A will satisfy the curve......

X⁴+ax²y+bxy+cx+dy+6=0 

(r/2)^4 + a(r/2)^2.(r)(root(3)/2 + b.root(3).r^2/4 + c.r/2 + d.root(3).r/2 + 6 =0 

r ^4 + 2.k.r^3 + 4.l.r^2 + 8.c.r + 8root(3).d.r + 96 = 0

Here we get equation in variable r....where r is the distance OA.....

This equation will give four roots of r...and that four root will be of course OA,OB,OC,OD

so OA,OB,OC,OD = product of roots of the above final equation = 96  ans...

is the answer 96. if it is then tell me i will post the complete solution

i have checked that the answer posted by the forrumn expert and my answer is correct but i think i have an easier method

first substitute the vallue of y=3^1/2 x to get a byquadratic eqn in x as

x^4+ 3^1/2 a x^3 + 3^1/2 b x^2 + (3^1/2 d +c) x+6 = 0

now, the product of the roots of the equation x1 x2 x3 x4 = 6

distance of A from origin ( x1^2 + y1^2)^1/2 now, since x1, y1 lie on y = 3^1/2 x so y1 = 3^1/2 x1

this implies OA = (x1^2 +3 x1^2)^1/2= 2x1

similarly OB = 2 x2 OC = 2 x3 OD = 2 x4

so OA.OB.OC.OD = 16 x1 x2 x3 x4 =16 x 6 = 96

did you get it!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!