Locus of z such that -z-z1-z-z2- constant is-

The correct option is D HyperbolaDifference of distances of a point from two fixed points nbsp; z_1 nbsp;and nbsp; z_2 nbsp;is constant.Thus ...

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Centre of Hyperbola

Locus of z ...

Question

Locus of $z$ such that $| z - z_{1} | - | z - z_{2} | = λ$ (constant) is:

Circle

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Parabola

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Ellipse

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Hyperbola

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Solution

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| z - z_{1} | + | z - z_{2} | = λ

∣ ∣ \frac{z + 1}{z - 1} ∣ ∣ = 1

∣ ∣ \frac{z - z_{1}}{z - z_{2}} ∣ ∣ = 1

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z_{1} = 10 + 6 i, z_{2} = 4 + 2 i

arg (\frac{z - z_{1}}{z - z_{2}}) = \frac{π}{4},

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z_{2}

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\frac{z - z_{1}}{z - z_{2}} = e^{i θ}, θ \in R

If $| z - z_{1} | + | z + z_{2} |$ = $λ, λ \in R^{+}, z_{1}, z_{2}$ are fixed complex numbers, represents an ellipse, then:

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