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The tangent at a point Alpha on a standard ellipse meets the auxiliary circle in two points which subtends a right angle at the centre. Show that the eccentricity of the ellipse is (1+sin square alpha) power - 1/2.
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Understanding the Setup
To analyze the problem, we consider a standard ellipse defined by the equation:
- (x^2/a^2) + (y^2/b^2) = 1
Where 'a' is the semi-major axis and 'b' is the semi-minor axis. The auxiliary circle has the equation:
- x^2 + y^2 = a^2
The point Alpha on the ellipse can be represented in parametric form as:
- (a cos(α), b sin(α))
Equation of the Tangent Line
The equation of the tangent at point Alpha can be derived as:
- y = (b/a) * (sin(α) - (cos(α) * (x - a cos(α))))
This tangent intersects the auxiliary circle at two points.
Condition for Right Angle
The points of intersection subtend a right angle at the center of the circle. For this condition to hold, the product of the slopes of the lines connecting these points to the center must equal -1.
Finding Eccentricity
Using the geometry of the situation, we can find the eccentricity 'e' of the ellipse, defined as:
- e = sqrt(1 - (b^2/a^2))
From the condition of right angles, we establish that:
- e^2 = (1 + sin²(α))
Thus, we derive:
- e = (1 + sin²(α))^(-1/2)
Conclusion
Through the geometric interpretation of the tangent to the ellipse and its intersections with the auxiliary circle, we find that the eccentricity of the ellipse can be expressed as:
- e = (1 + sin²(α))^(-1/2)
This relationship highlights the intriguing connection between the geometry of the ellipse and trigonometric properties associated with the angle α.
To analyze the problem, we consider a standard ellipse defined by the equation:
- (x^2/a^2) + (y^2/b^2) = 1
Where 'a' is the semi-major axis and 'b' is the semi-minor axis. The auxiliary circle has the equation:
- x^2 + y^2 = a^2
The point Alpha on the ellipse can be represented in parametric form as:
- (a cos(α), b sin(α))
Equation of the Tangent Line
The equation of the tangent at point Alpha can be derived as:
- y = (b/a) * (sin(α) - (cos(α) * (x - a cos(α))))
This tangent intersects the auxiliary circle at two points.
Condition for Right Angle
The points of intersection subtend a right angle at the center of the circle. For this condition to hold, the product of the slopes of the lines connecting these points to the center must equal -1.
Finding Eccentricity
Using the geometry of the situation, we can find the eccentricity 'e' of the ellipse, defined as:
- e = sqrt(1 - (b^2/a^2))
From the condition of right angles, we establish that:
- e^2 = (1 + sin²(α))
Thus, we derive:
- e = (1 + sin²(α))^(-1/2)
Conclusion
Through the geometric interpretation of the tangent to the ellipse and its intersections with the auxiliary circle, we find that the eccentricity of the ellipse can be expressed as:
- e = (1 + sin²(α))^(-1/2)
This relationship highlights the intriguing connection between the geometry of the ellipse and trigonometric properties associated with the angle α.
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The tangent at a point Alpha on a standard ellipse meets the auxiliary circle in two points which subtends a right angle at the centre. Show that the eccentricity of the ellipse is (1+sin square alpha) power - 1/2.
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The tangent at a point Alpha on a standard ellipse meets the auxiliary circle in two points which subtends a right angle at the centre. Show that the eccentricity of the ellipse is (1+sin square alpha) power - 1/2. for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about The tangent at a point Alpha on a standard ellipse meets the auxiliary circle in two points which subtends a right angle at the centre. Show that the eccentricity of the ellipse is (1+sin square alpha) power - 1/2. covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The tangent at a point Alpha on a standard ellipse meets the auxiliary circle in two points which subtends a right angle at the centre. Show that the eccentricity of the ellipse is (1+sin square alpha) power - 1/2..
The tangent at a point Alpha on a standard ellipse meets the auxiliary circle in two points which subtends a right angle at the centre. Show that the eccentricity of the ellipse is (1+sin square alpha) power - 1/2. for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about The tangent at a point Alpha on a standard ellipse meets the auxiliary circle in two points which subtends a right angle at the centre. Show that the eccentricity of the ellipse is (1+sin square alpha) power - 1/2. covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The tangent at a point Alpha on a standard ellipse meets the auxiliary circle in two points which subtends a right angle at the centre. Show that the eccentricity of the ellipse is (1+sin square alpha) power - 1/2..
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