Tangent is drawn at any point P of a curve which passes through (1, 1) cutting x-axis and y-axis at A and B respectively. If AP:BP = 3:1, then ,

Options

A. the differential equation of the curve is `3x(dy)/(dx)+y=0` and the curve passes through `(1//8,2)`
B. the differential eqaution of the curve is `3x(dy)/(dx)-y=0` and the curve pass through `(1//8,-2)`
C. the curve is passing through `(-1//8,-2)`
D. the normal at (1,1) is `x+3y=4`

Answer

A

Explanation

The eqaution of the tangent at `P(x,y)` is `Y-y=(dy)/(dx)(X-x)`
It cuts the coordinate axes at

It is given that `AP:BP=3:1`
`therefore" "(3xx0+x-y(dx)/(dy))/(3+1)=x and ((y-x(dy)/(dx))+1xx0)/(3+1)=y`
`rArr" "x-y(dx)/(dy)=4x and 3y-3x(dy)/(dx)=4y`
`rArr" "(dy)/(dx)=-(y)/(3x)rArr3x(dy)/(dx)+y=0rArr(3)/(y)dy+(1)/(x)dx=0`
Integrating, we get
`y^(3)x=C" ...(i)"`
It passes through (1, 1)
`therefore" "C=1`
Putting C = 1 in (i), we obtain `y^(3)x=1` as the equation of the curve. Clearly, it passes through (1/8,2).