Question:
The locus of \( |z-2i|+|z+4i|=10 \) is
The locus of \( |z-2i|+|z+4i|=10 \) is
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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 \( z_2 \).
• 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 \( z_2 \).
Updated On: Oct 16, 2026
- a circle with centre at (0,-1) and radius 5
- a parabola with focus at (0,-1)
- a hyperbola with foci at (0, 2) and (0,-4)
- an ellipse with eccentricity \( 3/5 \)
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The Correct Option is D
Solution and Explanation
Concept:
In the complex plane, an equation of the form \( |z - z_1| + |z - z_2| = 2a \) represents an ellipse with foci at points \( z_1 \) and \( z_2 \), provided that the constant length of the major axis \( 2a \) is strictly greater than the distance between the two foci \( |z_1 - z_2| \).
The distance between the foci is given by \( 2ae = |z_1 - z_2| \), where \( e \) is the eccentricity.
Step 1: Identifying the focus points and constants.
Let us rewrite the given expression to fit the standard form: \[ |z - 2i| + |z - (-4i)| = 10 \] Comparing this to the standard ellipse formula:
• Foci: \( z_1 = 2i \equiv (0, 2) \) and \( z_2 = -4i \equiv (0, -4) \)
• Major axis length: \( 2a = 10 \implies a = 5 \)
Step 2: Calculating the distance between the foci and verifying consistency.
The distance between the focal points is: \[ |z_1 - z_2| = |2i - (-4i)| = |6i| = 6 \] Since \( 2a = 10 > 6 \), the condition for an ellipse is perfectly satisfied.
Step 3: Finding the eccentricity \( e \).
The focal distance formula states: \[ 2ae = |z_1 - z_2| \implies 2(5)e = 6 \implies 10e = 6 \implies e = \frac{6}{10} = \frac{3}{5} \] Thus, the locus is an ellipse with an eccentricity value of \( 3/5 \).
Step 1: Identifying the focus points and constants.
Let us rewrite the given expression to fit the standard form: \[ |z - 2i| + |z - (-4i)| = 10 \] Comparing this to the standard ellipse formula:
• Foci: \( z_1 = 2i \equiv (0, 2) \) and \( z_2 = -4i \equiv (0, -4) \)
• Major axis length: \( 2a = 10 \implies a = 5 \)
Step 2: Calculating the distance between the foci and verifying consistency.
The distance between the focal points is: \[ |z_1 - z_2| = |2i - (-4i)| = |6i| = 6 \] Since \( 2a = 10 > 6 \), the condition for an ellipse is perfectly satisfied.
Step 3: Finding the eccentricity \( e \).
The focal distance formula states: \[ 2ae = |z_1 - z_2| \implies 2(5)e = 6 \implies 10e = 6 \implies e = \frac{6}{10} = \frac{3}{5} \] Thus, the locus is an ellipse with an eccentricity value of \( 3/5 \).
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