`bb"statement-1"` Locus of z satisfying the equation `abs(z-1)+abs(z-8)=5` is an ellipse.
`bb"statement-2"` Sum of focal distances of any point on ellipse is constant for an ellipse.

locus of the point z satisfying the equation |iz-1|+|z-i|=2 is

Statement 1 : In an ellipse, the sum of the distances between foci is always less than the sum of focal distances of any point on it. Statement 2 : The eccentricity of any ellipse is less than 1.

The solution of the equation |z|-z=1+2i is :

Solve the equation |z|=z+1+2idot

The equation (x^(2))/(10-a)+(y^(2))/(4-a)=1 represents an ellipse , if

Consider the curves on the Argand plane as " "{:(C_(1):arg(z)=pi/4","),(C_(2):arg(z)=(3pi)/(4)):} and C_(3):arg(z-5-5i)=pi," where " i=sqrt(-1). bb"Statement-1" Area of the region bounded by the curves C_(1),C_(2) " and " C_(3) " is " 25/2 bb"Statement-2" The boundaries of C_(1),C_(2) " and " C_(3) constitute a right isosceles triangle.

The complex number z = x + iy , which satisfies the equation |(z-5i)/(z+5i)| =1 , lies on:

The eccentric angle of a point on the ellipse x^(2)/6 + y^(2)/2 = 1 whose distance from the centre of the ellipse is 2, is

If z=x+iy, where i=sqrt(-1) , then the equation abs(((2z-i)/(z+1)))=m represents a circle, then m can be

Show that the equation (10x-5)^(2)+(10y-5)^(2)=(3x+4y-1)^(2) represents an ellipse, find the eccentricity of the ellipse.