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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 xdydx3y=0
C curve passes through (2, 1/8)
Given BP:AP=3:1. Then equation of tangent is

Yy=f(x)(Xx)

The intercept on the coordinate axes are

A(xyf(x)=0) and B[0,yxf(x)]

Since,P is internally intercepts a line AB,

[x=(mx1+nx2m+n)] by using this formula

x=3(xyf(x))+1×03+1 dydx=y3x dyy=13xdx

On integrating both sides , we get

xy3=C

Since, curve passes through (1,1),thenc=1

xy3=1

At x=18y=2

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