For any value of , if the straight lines x +(1- ) y=a and x -(1+ ) y+a =0 intersect at P( ) , then the locus of P( ) is a

Step-by-step solution

Given, equations of straight lines are Subtracting Eq. (ii) from Eq. (i), we get array ll & (1- ) y+(1+ ) y=2 a & y[1- +1+ ]=2 a & y=a array Putting the value of y in Eq. (i), we get array ll & x +(1- ) a =a & [x+(1- ) a]=a & x+a-a =a & x-a =0 x=a Now, & x^2+y^2=(a )^2+(a )^2 & x^2+y^2=a^2 ^2 +a^2 ^2 & x^2+y^2=a^2 ( ^2 + ^2 ) [ ^2 + ^2 =1 ] & x^2+y^2=a^2, whose represent a circle. array