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# A tangent to the curve y=f(x) atP(x,y) cuts the x-axis and the y-axis at A and B, respectively, such thatBP:AP=3:1. Iff(1)=1 then :

A
the equation of the curve is xdydx+3y=0
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B
the curve passes through (2,18)
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C
the equation of the curve is xdydx3y=0
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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 equation of the curve is xdydx+3y=0 C the curve passes through (2,18)Equation of tangent is Y−y=dydx(X−x)Given BPAP=31 so that⇒dxx=−dy3y⇒xdydx+3y=0⇒tanx3=−tancy⇒1x3=cyGiven f(1)=1⇒c=1∴y=1x2and point (2,18) satisfy this.   Suggest Corrections  0      Similar questions  Related Videos   Methods of Solving First Order, First Degree Differential Equations
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