If , are the two real roots of 4^ tn roots of unity and , are the other two roots of it, then the sum of the eccentricities of the conics |z- |+|z- |=4 and |z- |+|z- |=6 is

Step-by-step solution

The real solutions of 4^ th root of unity is 1 =1, =-1 and other roots are i aligned & =i & =-i & |z-1|+|z+1|=4 & |x+i y-1|+|x-i y+1|=4 & |(x-1)+i y|+|(x+1)-i y|=4 & (x-1)^2+y^2 =4- (x+1)^2+y^2 & (x-1)^2+y^2=16+(x+1)^2+y^2-8 (x+1)^2+y^2 & -4 x=16-8 (x+1)^2+y^2 & 2 (x+1)^2+y^2 =4+x aligned aligned & 4(x+1)^2+4 y^2=(4+x)^2 & x^2 4 + y^2 3 =1 aligned which is an ellipse whose eccentricity is e₁= 1- 3 4 = 1 2 Now, |z- |+|z+ |=6 |z-i|+|z+i|=6 aligned & |x+i(y-1)|+|x+i(y+1)|=6 & x^2+(y-1)^2 + x^2(y+1)^2 =6 aligned x^2+(y