`O A` and `O B` are fixed straight lines, `P` is any point and `P M` and `P N` are the perpendiculars from `P` on `O Aa n dO B ,` respectively. Find the locus of `P` if the quadrilateral `O M P N` is of constant area.

Explanation

Taking OA as the x-axis, O as the origin, let the equation of OB be y = mx.
Then
`M-=(x_(1),0)`
The equation of hte perpendicular PN is
`my+x=my_(1)+x_(1)`
Solving the equations of OB and PN, we get
`N-=((my_(1)+x_(1))/(1+m^(2)),(m(my_(1)+x_(1)))/(1+x^(2)))`
`therefore" Area of quadrilateral OMPN (by stair method)"`
`=(1)/(2)|(0,0),(x_(1),0),(x_(1),y_(1)),((my_(1)+x_(1))//(1+x^(2)),m(my_(1)+x_(1))//(1+m^(2))),(0,0)|`
`[(1)/(2)x_(1)y_(1)+x_(1)(m(my_(1)+x_(1)))/(1+x^(2))-y_(1)(my_(1)+x_(1))/(1+m^(2))]`
`=pmk("say")`
Therefore, the locus P is
`mx^(2)+2m^(2)xy-my^(2)pm2(1+m^(2))k=0`
Here, `h=x^(2)`
a = m
`b=-m`
`f=g=0`
`"and "c=pm2(1+m^(2))k`
`therefore" "Delta=abc+2fgh-af^(2)-bg^(2)-ch^(2)`
`=pm2m^(2)(1+m^(2))kpm2m^(4)(1+m^(2))k^(2) ne0`
`"and "h^(2)gtab`
So, the locus is a hyperbola.