Chapter 18: Problem 24
If two points \(A(a, 0)\) and \(B(-a, 0)\) are stationary and if \(\angle A-\angle B=\theta\) in \(\Delta A B C\), the locus of \(C\) is (A) \(x^{2}+y^{2}+2 x y \tan \theta=a^{2}\) (B) \(x^{2}-y^{2}+2 x y \tan \theta=a^{2}\) (C) \(x^{2}+y^{2}+2 x y \cot \theta=a^{2}\) (D) \(x^{2}-y^{2}+2 x y \cot \theta=a^{2}\)
Short Answer
Step by step solution
Understanding the Problem
Using Angle Condition
Setting Up the Cosine Rule
Simplifying the Expression
Verifying the Locus Equation
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Geometry and Locus
The theoretical foundation lies in understanding that these points form a triangle. By defining the relationship among these points, we can find an equation that characterizes the locus. The solution utilizes the relationship that involves angles and distances between these points to form a practical equation of locus, showcasing the power of geometry not just to position points, but to describe paths and motion.
Trigonometry in Triangle Analysis
Initially, the cosine rule gives:
- \( \cos \angle ACB = \frac{\overrightarrow{CA} \cdot \overrightarrow{CB}}{|\overrightarrow{CA}| \times |\overrightarrow{CB}|} \)
Through algebraic manipulation and the use of trigonometric identities (such as \( \tan \theta = \frac{\sin \theta}{\cos \theta} \)), we unravel the locus condition: \( x^2 + y^2 + 2xy \tan \theta = a^2 \).
Coordinate Geometry Applications
The vectors \( \overrightarrow{CA} \) and \( \overrightarrow{CB} \) simplify the visualization and computation of distances and angles. By dealing in coordinates:
- \( \overrightarrow{CA} = (x-a, y) \)
- \( \overrightarrow{CB} = (x+a, y) \)
This methodical approach demonstrates how geometry can extend beyond shape and form, giving us powerful tools to solve complex problems using just positions expressed numerically.