As the graph of the equation x^3+3y^3= 4xy
shows, each line y = tx intersects the graph
at the origin and one other point. So every
point P = (0 6 , 0) on the graph can be written
parametrically as
P = (x(t), y(t)).
find y(t)
*answer choices*:
1. y(t) =4/1 + 3t^3
2. y(t) =4t^2/3 + t^3
3. y(t) =4t^2/1 + 3t^3
4. y(t) =4/3 + t^3
5. y(t) =4t/3 + t^3
6. y(t) =4t/1 + 3t^3

Question

As the graph of the equation x^3+3y^3= 4xy
shows, each line y = tx intersects the graph
at the origin and one other point. So every
point P = (0 6 , 0) on the graph can be written
parametrically as
P = (x(t), y(t)).
find y(t)
*answer choices*:
1. y(t) =4/1 + 3t^3
2. y(t) =4t^2/3 + t^3
3. y(t) =4t^2/1 + 3t^3
4. y(t) =4/3 + t^3
5. y(t) =4t/3 + t^3
6. y(t) =4t/1 + 3t^3

dumbcow · Answer

i think you substitute x with y/t

--> y^3/t^3 +3y^3 = 4y^2/t

--> (3+1/t^3)y^3 - (4/t)y^2 = 0
divide by y^2
--> (3+1/t^3)y - 4/t = 0

--> y = 4/t / (3+1/t^3)

--> y = 4t^2/(3t^3 +1)