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Question
Mirror image of the curve ∣∣∣z−z1z−z2∣∣∣=a, aϵR+, a≠1, about the line |z−z1|=|z−z2| is given by
Mirror image of the curve ∣∣∣z−z1z−z2∣∣∣=a, aϵR+, a≠1, about the line |z−z1|=|z−z2| is given by
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Solution
The correct option is D ∣∣∣z−z2z−z1∣∣∣=a
∣∣∣z−z1z−z2∣∣∣=a, line |z−z1|=|z−z2|We can assume the local of the final image to be (h,k) of the image of (h,k) on the given line, the image of the locus lie on the original curve.
⇒∣∣∣z−z2z−z1∣∣∣=aHence, the answer is ∣∣∣z−z2z−z1∣∣∣=a.
∣∣∣z−z1z−z2∣∣∣=a, line |z−z1|=|z−z2|
We can assume the local of the final image to be (h,k) of the image of (h,k) on the given line, the image of the locus lie on the original curve.
⇒∣∣∣z−z2z−z1∣∣∣=a
Hence, the answer is ∣∣∣z−z2z−z1∣∣∣=a.
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