JEE Advance - Mathematics (2006 - No. 6)
A curve $$y=f(x)$$ passes through $$(1,1)$$ and at $$P(x,y),$$ tangent cuts the $$x$$-axis and $$y$$-axis at $$A$$ and $$B$$ respectively such that $$BP:AP=3:1,$$ then
equation of curve is $$xy'-3y=0$$
normal at $$(1,1)$$ is $$x+3y=4$$
curve passes through $$(2, 1/8)$$
equation of curve is $$xy'+3y=0$$