Question

# A tangent drawn to the curve y=f(x) at P(x,y)cuts the x-axis and y-axis at A and B respectively such that BP: AP =3 : 1, given that f(1)=1, then

A
equation of curve is xdydx3y=0
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B
normal at (1, 1) is x+3y=4
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C
curve passes through (2, 1/8)
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D
equation of curve is xdydx+3y=0
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Solution

## The correct options are A equation of curve is xdydx−3y=0 C curve passes through (2, 1/8)Given BP:AP=3:1. Then equation of tangent is Y−y=f′(x)(X−x)The intercept on the coordinate axes are A(x−yf′(x)=0) and B[0,y−xf′(x)]Since,P is internally intercepts a line AB,[x=(mx1+nx2m+n)] by using this formula ∴x=3(x−yf′(x))+1×03+1 ⇒dydx=y−3x ⇒dyy=−13xdxOn integrating both sides , we get xy3=C Since, curve passes through (1,1),thenc=1 ∴xy3=1 ∴ At x=18⇒y=2

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