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The eccentricity of an ellipse, with its centre at the origin, is . If one of the directrices is x = 4, then the equation of the ellipse is- [AIEEE 2004]
- a)3x2 + 4y2 =1
- b)3x2 + 4y2 =12
- c)4x2 + 3y2 =12
- d)4x2 + 3y2 =1
Correct answer is option 'B'. Can you explain this answer?
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The eccentricity of an ellipse, with its centre at the origin, is . If...
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The eccentricity of an ellipse, with its centre at the origin, is . If...
Understanding the Eccentricity of the Ellipse
The eccentricity (e) of an ellipse is defined as the ratio of the distance from a point on the ellipse to the focus, over the distance to the corresponding directrix. For an ellipse centered at the origin, the relationship between the semi-major axis (a), semi-minor axis (b), and eccentricity (e) is given by:
- e = sqrt(1 - (b^2/a^2))
Given that one of the directrices is x = 4, we can identify the ellipse's properties.
Identifying the Parameters
- The distance from the center to the directrix (d) is 4.
- For an ellipse, the directrix corresponding to the semi-major axis a is given by the formula: x = a/e.
- Thus, we can express: a/e = 4.
Setting up the Equation
From the relationship a = 4e, we can substitute this into the eccentricity formula:
- e = 4e^2 + b^2 = 1
- Rearranging gives us: b^2 = a^2(1 - e^2).
We know e < 1,="" which="" implies="" b^2="a^2(1" -="" e^2)="" will="" yield="" a="" positive="" value="" for="" />
Finalizing the Equation
Considering the standard form of an ellipse centered at the origin:
- (x^2/a^2) + (y^2/b^2) = 1.
After substituting a and b based on the calculated values, we find:
- For a = 4, b^2 = 12.
- Therefore, we have: 4x^2 + 3y^2 = 12.
Conclusion
After simplifying, we arrive at the final equation of the ellipse:
- The correct answer is option b) 3x^2 + 4y^2 = 12.
The eccentricity (e) of an ellipse is defined as the ratio of the distance from a point on the ellipse to the focus, over the distance to the corresponding directrix. For an ellipse centered at the origin, the relationship between the semi-major axis (a), semi-minor axis (b), and eccentricity (e) is given by:
- e = sqrt(1 - (b^2/a^2))
Given that one of the directrices is x = 4, we can identify the ellipse's properties.
Identifying the Parameters
- The distance from the center to the directrix (d) is 4.
- For an ellipse, the directrix corresponding to the semi-major axis a is given by the formula: x = a/e.
- Thus, we can express: a/e = 4.
Setting up the Equation
From the relationship a = 4e, we can substitute this into the eccentricity formula:
- e = 4e^2 + b^2 = 1
- Rearranging gives us: b^2 = a^2(1 - e^2).
We know e < 1,="" which="" implies="" b^2="a^2(1" -="" e^2)="" will="" yield="" a="" positive="" value="" for="" />
Finalizing the Equation
Considering the standard form of an ellipse centered at the origin:
- (x^2/a^2) + (y^2/b^2) = 1.
After substituting a and b based on the calculated values, we find:
- For a = 4, b^2 = 12.
- Therefore, we have: 4x^2 + 3y^2 = 12.
Conclusion
After simplifying, we arrive at the final equation of the ellipse:
- The correct answer is option b) 3x^2 + 4y^2 = 12.
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The eccentricity of an ellipse, with its centre at the origin, is . If one of the directrices isx = 4, then the equation of the ellipse is- [AIEEE 2004]a)3x2+ 4y2=1b)3x2+ 4y2=12c)4x2+ 3y2=12d)4x2+ 3y2=1Correct answer is option 'B'. Can you explain this answer?
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The eccentricity of an ellipse, with its centre at the origin, is . If one of the directrices isx = 4, then the equation of the ellipse is- [AIEEE 2004]a)3x2+ 4y2=1b)3x2+ 4y2=12c)4x2+ 3y2=12d)4x2+ 3y2=1Correct answer is option 'B'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The eccentricity of an ellipse, with its centre at the origin, is . If one of the directrices isx = 4, then the equation of the ellipse is- [AIEEE 2004]a)3x2+ 4y2=1b)3x2+ 4y2=12c)4x2+ 3y2=12d)4x2+ 3y2=1Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The eccentricity of an ellipse, with its centre at the origin, is . If one of the directrices isx = 4, then the equation of the ellipse is- [AIEEE 2004]a)3x2+ 4y2=1b)3x2+ 4y2=12c)4x2+ 3y2=12d)4x2+ 3y2=1Correct answer is option 'B'. Can you explain this answer?.
The eccentricity of an ellipse, with its centre at the origin, is . If one of the directrices isx = 4, then the equation of the ellipse is- [AIEEE 2004]a)3x2+ 4y2=1b)3x2+ 4y2=12c)4x2+ 3y2=12d)4x2+ 3y2=1Correct answer is option 'B'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The eccentricity of an ellipse, with its centre at the origin, is . If one of the directrices isx = 4, then the equation of the ellipse is- [AIEEE 2004]a)3x2+ 4y2=1b)3x2+ 4y2=12c)4x2+ 3y2=12d)4x2+ 3y2=1Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The eccentricity of an ellipse, with its centre at the origin, is . If one of the directrices isx = 4, then the equation of the ellipse is- [AIEEE 2004]a)3x2+ 4y2=1b)3x2+ 4y2=12c)4x2+ 3y2=12d)4x2+ 3y2=1Correct answer is option 'B'. Can you explain this answer?.
Solutions for The eccentricity of an ellipse, with its centre at the origin, is . If one of the directrices isx = 4, then the equation of the ellipse is- [AIEEE 2004]a)3x2+ 4y2=1b)3x2+ 4y2=12c)4x2+ 3y2=12d)4x2+ 3y2=1Correct answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
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