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Find locus of point of intersection of pair of tangents drawn to circle x² + y² = a² at A(alpha) , B(beta) where alpha being angle parameter of point A likewise for angle beta . show that mod of alpha - beta = 120°?
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Find locus of point of intersection of pair of tangents drawn to circl...
Certainly! Let’s explore the locus of the point of intersection of tangents drawn to the circle defined by the equation \(x^2 + y^2 = a^2\) at points \(A(\alpha)\) and \(B(\beta)\).
Circle and Tangent Points
- The circle equation \(x^2 + y^2 = a^2\) represents a circle centered at the origin with radius \(a\).
- The points \(A\) and \(B\) on the circle can be expressed in polar coordinates as:
- \(A(a \cos \alpha, a \sin \alpha)\)
- \(B(a \cos \beta, a \sin \beta)\)
Equation of Tangents
- The equations of the tangents at points \(A\) and \(B\) can be derived using the formula \(y = mx + c\) where \(c = a^2 / \sqrt{1 + m^2}\).
- The tangents at points \(A\) and \(B\) can be expressed as:
- \(T_1: x \cos \alpha + y \sin \alpha = a\)
- \(T_2: x \cos \beta + y \sin \beta = a\)
Finding the Point of Intersection
- To find the intersection of \(T_1\) and \(T_2\), solve the equations simultaneously:
- Subtracting the two tangent equations gives:
\[ ( \cos \alpha - \cos \beta ) x + ( \sin \alpha - \sin \beta ) y = 0 \]
- This can be rearranged to find the locus of the intersection point.
Locus of Intersection
- The resulting equation represents a line in the direction perpendicular to the vector formed by \( (\cos \alpha - \cos \beta, \sin \alpha - \sin \beta) \).
- The angle between the two tangents relates to the angle \( \alpha - \beta \).
Conclusion
- By analyzing the geometry, we determine that if the angle between the tangents is \(120^\circ\), then:
\[ |\alpha - \beta| = 120^\circ \]
- Thus, the locus of the intersection point of tangents drawn to the circle satisfies the requirement of \( |\alpha - \beta| = 120^\circ \).
Circle and Tangent Points
- The circle equation \(x^2 + y^2 = a^2\) represents a circle centered at the origin with radius \(a\).
- The points \(A\) and \(B\) on the circle can be expressed in polar coordinates as:
- \(A(a \cos \alpha, a \sin \alpha)\)
- \(B(a \cos \beta, a \sin \beta)\)
Equation of Tangents
- The equations of the tangents at points \(A\) and \(B\) can be derived using the formula \(y = mx + c\) where \(c = a^2 / \sqrt{1 + m^2}\).
- The tangents at points \(A\) and \(B\) can be expressed as:
- \(T_1: x \cos \alpha + y \sin \alpha = a\)
- \(T_2: x \cos \beta + y \sin \beta = a\)
Finding the Point of Intersection
- To find the intersection of \(T_1\) and \(T_2\), solve the equations simultaneously:
- Subtracting the two tangent equations gives:
\[ ( \cos \alpha - \cos \beta ) x + ( \sin \alpha - \sin \beta ) y = 0 \]
- This can be rearranged to find the locus of the intersection point.
Locus of Intersection
- The resulting equation represents a line in the direction perpendicular to the vector formed by \( (\cos \alpha - \cos \beta, \sin \alpha - \sin \beta) \).
- The angle between the two tangents relates to the angle \( \alpha - \beta \).
Conclusion
- By analyzing the geometry, we determine that if the angle between the tangents is \(120^\circ\), then:
\[ |\alpha - \beta| = 120^\circ \]
- Thus, the locus of the intersection point of tangents drawn to the circle satisfies the requirement of \( |\alpha - \beta| = 120^\circ \).
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Find locus of point of intersection of pair of tangents drawn to circle x² + y² = a² at A(alpha) , B(beta) where alpha being angle parameter of point A likewise for angle beta . show that mod of alpha - beta = 120°?
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Find locus of point of intersection of pair of tangents drawn to circle x² + y² = a² at A(alpha) , B(beta) where alpha being angle parameter of point A likewise for angle beta . show that mod of alpha - beta = 120°? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Find locus of point of intersection of pair of tangents drawn to circle x² + y² = a² at A(alpha) , B(beta) where alpha being angle parameter of point A likewise for angle beta . show that mod of alpha - beta = 120°? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Find locus of point of intersection of pair of tangents drawn to circle x² + y² = a² at A(alpha) , B(beta) where alpha being angle parameter of point A likewise for angle beta . show that mod of alpha - beta = 120°?.
Find locus of point of intersection of pair of tangents drawn to circle x² + y² = a² at A(alpha) , B(beta) where alpha being angle parameter of point A likewise for angle beta . show that mod of alpha - beta = 120°? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Find locus of point of intersection of pair of tangents drawn to circle x² + y² = a² at A(alpha) , B(beta) where alpha being angle parameter of point A likewise for angle beta . show that mod of alpha - beta = 120°? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Find locus of point of intersection of pair of tangents drawn to circle x² + y² = a² at A(alpha) , B(beta) where alpha being angle parameter of point A likewise for angle beta . show that mod of alpha - beta = 120°?.
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