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Question

A curve y=f(x) passes through (1,1) and tangent at P(x,y) cuts the xaxis and yaxis at A and B respectively such that AP:PB=1:3, then which of the following is/are true:

A
Differential equation of curve is xy3y=0
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B
Differential equation of curve is xy+3y=0
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C
Curve passes through (2,18)
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D
Curve passes through (2,14)
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Solution

The correct option is C Curve passes through (2,18)
Tangent to the curve y=f(x) at P(x,y) is
Yy=dydx(Xx)
A⎜ ⎜ ⎜xdydxydydx,0⎟ ⎟ ⎟;B(0,xdydx+y)
AP:PB=1:3
Using section formula, we have:
x=3(xdydxy)dydx+1×04
xdydx+3y=0
( Note that the same we will get with yco-ordinate)
Which can also be written as xy+3y=0
dyy=3dxx
ln|y|=3ln|x|+ln|c|
y=cx3
As curve passes through (1,1),c=1
curve is x3y=1 which also passes through (2,18)

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