The solution of the equation `(2+x)dy-ydx=0` represents a curve passing through a fixed point P, then the area of equilateral triangle with P as one vertex and `x+y=0` as its one side, is

Options

A. `2sqrt3`
B. `sqrt3`
C. `(2)/(sqrt3)`
D. `(4)/(sqrt3)`

Answer

C

Explanation

We have,
`(2+x)dy-ydx=0rArr(1)/(y)dx-(1)/(2+x)dx=0`
On integrating, we obtain
`logy-log(x+2)=logC rArr y=C(x+2)`
Clearly, it represents a curve passing through a fixed point `P(-2,0)`.
The altitude of the equilateral triangle having one vertex at `P(-2,0)` and opposite side `x+y=0` is
`P=|(-2+0)/(sqrt(1+1))|=sqrt2`
So, its area is `(p^(2))/(sqrt3)=(2)/(sqrt3)` sq. units