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Show that the curve such that the ratio of the di stance between the normal at any of its points and the origin to the distance between the sam e normal and the point (a, b) is equal to the constant k (k > 0, k = 1)is a circle.

Show that the curve such that the ratio of the di stance between the normal at any of its points and the origin to the distance between the sam e normal and the point (a, b) is equal to the constant k (k > 0, k = 1)is a circle.

Algebra and Trigonometry (6th Edition)
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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: Show that the curve such that the ratio of the di stance between the normal at any of its points and the origin to the distance between the sam e normal and the point (a, b) is equal to the constant k (k> 0, k +1)is a circle.
Transcribed Image Text:: Show that the curve such that the ratio of the di stance between the normal at any of its points and the origin to the distance between the sam e normal and the point (a, b) is equal to the constant k (k> 0, k +1)is a circle.
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