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# A curve y=f(x) passes through (1,1) and tangent at P(x,y) cuts the x−axis and y−axis 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 Y−y=dydx(X−x) ∴A⎛⎜ ⎜ ⎜⎝xdydx−ydydx,0⎞⎟ ⎟ ⎟⎠;B(0,−xdydx+y) ∵AP:PB=1:3 Using section formula, we have: x=3(xdydx−y)dydx+1×04 ⇒xdydx+3y=0 ( Note that the same we will get with y−co-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)  Suggest Corrections  0      Similar questions  Explore more