The combined equation of bisectors of lines `y = m_(1) x " and " y=m_(2)x " if "m _(1),m_(2)` are roots of the equation `x^(2)-ax+1=0` is :

Options

A. `x^(2)-3y^(2)=0`
B. `x^(2)-y^(2)=0`
C. `xy=0`
D. `x^(2)-4xy=0`

Answer

C

Explanation

`because " "(dy)/(dx)=x+xyimplies(dy)/(dx)-xy=x`
`therefore " "l.F.=e^(intpdx)=e^(int-xdx)=e^(-(x^(2))/(2))`
`therefore` Solution of equation is `y.e^(-(x^(2))/(2))=intxe^(-(x^(2))/(2))dximpliesy.e^(-(x^(2))/(2))=-e^(-(x^(2))/(2))+c`
`because` It passes through `(0,0)impliesc=1impliesy.e^(-x^(2)/(2))=-e^(-x^(2)/(2))+1impliesy=e^(x^(2)/(2))-1`
`therefore " "f(x)=1impliese^(x^(2)/(2))-1=1impliese^(x^(2)/(2))=2impliesx^(2)=2ln2impliesx=pmsqrt(2ln2)`