Tangent is drawn at any pt. P of a curve which passes thru (1,1) cuttin x-axis & y-axis at A & B resp. If AP:BP = 3:1 then

a. differential eqn of the curve is 3xdy/dx + y =0

b. differential eqn of the curve is 3xdy/dx - y =0

c. curve is passing through (1/8,2)

d. normal at (1,1) is x+3y=4

 its a multiple choice quesn.

plz gimme detailed soln