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Question

Tangent is drawn at any point P(x,y) on a curve, which passes through (1,1). The tangent cuts X-axis and Y-axis at A and B respectively. If AP:BP=3:1, then

A
the differential equation of the curve is 3xdydx+y=0
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B
the differential equation of the curve is 3xdydxy=0
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C
the curve passes through (18,2)
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D
the normal at (1,1) is x+3y=4
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Solution

The correct options are
A the differential equation of the curve is 3xdydx+y=0
C the curve passes through (18,2)
Since BP:AP=3:1
Equation of the tangent is Yy=f(x)(Xx).
Intercept on X-axis (xyf(x)),0), Y-axis(0,yxf(x)).
Now x=(xyf(x))+1×03+1 {since, it divides internally 3:1}
dydx=y3xdyy=dx3x3xdydx+y=0
log y=13log x+log Cxy3=C
But curve passes through (1,1) 1=C
xy3=1
curve passes through (18,2).

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