Question

The locus of \( |z-2i|+|z+4i|=10 \) is

• A. a circle with centre at (0,-1) and radius 5
• B. a parabola with focus at (0,-1)
• C. a hyperbola with foci at (0, 2) and (0,-4)
• D. an ellipse with eccentricity \( 3/5 \)

Hint:

Remember these core complex locus rules:

• If \( |z - z_1| + |z - z_2| = k \) and \( k > |z_1 - z_2| \), the locus is an ellipse.

• If \( |z - z_1| - |z - z_2| = k \) and \( k < |z_1 - z_2| \), the locus is a hyperbola.

• If \( |z - z_1| = |z - z_2| \), the locus is the perpendicular bisector joining \( z_1 \) and \( z_2 \).

Answer 1

The Correct Option is D

Step 1: Read the equation as distances.
The term $|z-2i|$ is the distance from point $z$ to $2i$, and $|z+4i|$ is the distance to $-4i$. So the sum of two distances is fixed at 10.
Step 2: Recall the ellipse rule.
When the sum of distances from two fixed points stays constant, the path is an ellipse, as long as that sum is bigger than the gap between the two points.
Step 3: Name the two fixed points.
The foci are at $2i = (0,2)$ and $-4i = (0,-4)$. The fixed sum is $2a = 10$, so $a = 5$.
Step 4: Find the gap between foci.
The distance between them is $|2i-(-4i)| = |6i| = 6$.
Step 5: Check it is really an ellipse.
Since $10 > 6$, the ellipse condition holds.
Step 6: Find the eccentricity.
Use $2ae = $ distance between foci, so $2(5)e = 6$, giving $e = \dfrac{6}{10}$. \[ \boxed{\text{ellipse with } e = \tfrac{3}{5}} \]