The slope of tangent at (x,y) to a curve passing through (2, 1) is `(x^(2)+y^(2))/(2xy)`, then the equation of the curve is

Options

A. `2(x^(2)-y^(2))=3x`
B. `2(x^(2)-y^(2))=6y`
C. `x(x^(2)-y^(2))=6`
D. `x(x^(2)+y^(2))=10`

Answer

A

Explanation

According to the question.
`(dy)/(dx)=(x^(2)+y^(2))/(2xy)`
Put `y=vx`
`rArr (dy)/(dx)=v+x(dv)/(dx)`
`therefore v+x(dv)/(dx)=(x^(2)+v^(2)x^(2))/(2vx^(2))`
`rArr x (dv)/(dx)=((1+v^(2))/(2v)-v)`
`rArr (2v)/(1-v^(2))dv=(dx)/(x)`
On integrating both sides, we get
`-log(1-v^(2))=log x+log c`
`rArr -log(1-(1)/(4))=log x+log c " " ` ...(i)
This passes through (2, 1).
`therefore -log(1-(1)/(4))=log2+log c`
`rArr -log((3)/(4))=log 2c`
`rArr "log"(4)/(3)=log 2c rArr c=(2)/(3)`
On putting `c=(2)/(3)` in Eq. (i), we get
`log((x^(2))/(x^(2)-y^(2)))="log"(2)/(3)x`
`rArr 2(x^(2)-y^(2))=3x`