Let P(h,k) be any point on the given curve. Also let A(a,0) and B(0,b).
Since AP:PB=3:1
P(h,k)
(a/4, 3b/4)
Thus, a=4h and b=4k/3
So we can equate the slope of the tangent at P to the derivativative at P
y'(h,k)=(4k/3-0)/(0-4h)
y'(h,k)=-k/3h
3hy'(h,k)+k=0
Since P(h,k) is totally an arbitary point h and k can be replaced with x and y respectively.
Therefore the ODE of the given curve is:
3xy'+y=0
Hence the option (D).
Moderation Team
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