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An ellipse intersects the hyperbola 2x2 – 2y2 = 1 orthogonally.The eccentricity of the ellipse is reciprocal of that of the hyperbola. If the axes of the ellipse are along the coordinate axes, then (2009)
- a)equation of ellipse is x2 + 2y2 = 2
- b)the foci of ellipse are (±1, 0)
- c)equation of ellipse is x2 + 2y2 = 4
- d)the foci of ellipse are (±
, 0)
Correct answer is option 'A,B'. Can you explain this answer?
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An ellipse intersects the hyperbola 2x2 – 2y2 = 1 orthogonally.T...
The given hyperbola is
...(1)which is a rectangular hyperbola (i.e. a = b)
Let the ellipse be 
Its eccentricity = 
So, the equation of ellipse becomes x2 + 2 y2 = a2 ...(2)
Let the hyperbola (1) and ellipse (2) intersect each other at P( x1 ,y1).
Then slope of hyperbola (1) at P is given by
Let the hyperbola (1) and ellipse (2) intersect each other at P( x1 ,y1).
Then slope of hyperbola (1) at P is given by
and that of ellipse (2) at P is
As the two curves intersect orthogonally,
∴ m1m2 =-1
Also P ( x1,y1) lies on x2 -y2 = 
...(ii)Solving (i) and (ii), we get 
Also P ( x1,y1) lies on ellipse x2 + 2y2 = a2
1 + 1 = a2 or a2 =2∴ The required ellipse is x2 + 2y2 = 2 whose foci
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An ellipse intersects the hyperbola 2x2 – 2y2 = 1 orthogonally.T...
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An ellipse intersects the hyperbola 2x2 – 2y2 = 1 orthogonally.T...
The given hyperbola is
...(1)which is a rectangular hyperbola (i.e. a = b)
Let the ellipse be
Its eccentricity =
So, the equation of ellipse becomes x2 + 2 y2 = a2 ...(2)
Let the hyperbola (1) and ellipse (2) intersect each other at P( x1 ,y1).
Then slope of hyperbola (1) at P is given by
Let the hyperbola (1) and ellipse (2) intersect each other at P( x1 ,y1).
Then slope of hyperbola (1) at P is given by
and that of ellipse (2) at P is
As the two curves intersect orthogonally,
∴ m1m2 =-1
Also P ( x1,y1) lies on x2 -y2 =
...(ii)Solving (i) and (ii), we get
Also P ( x1,y1) lies on ellipse x2 + 2y2 = a2
1 + 1 = a2 or a2 =2∴ The required ellipse is x2 + 2y2 = 2 whose foci
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An ellipse intersects the hyperbola 2x2 – 2y2 = 1 orthogonally.The eccentricity of the ellipse is reciprocal of that of the hyperbola. If the axes of the ellipse are along the coordinate axes, then (2009)a)equation of ellipse is x2 + 2y2 = 2b)the foci of ellipse are (±1, 0)c)equation of ellipse is x2 + 2y2 = 4d)the foci of ellipse are (±, 0)Correct answer is option 'A,B'. Can you explain this answer?
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An ellipse intersects the hyperbola 2x2 – 2y2 = 1 orthogonally.The eccentricity of the ellipse is reciprocal of that of the hyperbola. If the axes of the ellipse are along the coordinate axes, then (2009)a)equation of ellipse is x2 + 2y2 = 2b)the foci of ellipse are (±1, 0)c)equation of ellipse is x2 + 2y2 = 4d)the foci of ellipse are (±, 0)Correct answer is option 'A,B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about An ellipse intersects the hyperbola 2x2 – 2y2 = 1 orthogonally.The eccentricity of the ellipse is reciprocal of that of the hyperbola. If the axes of the ellipse are along the coordinate axes, then (2009)a)equation of ellipse is x2 + 2y2 = 2b)the foci of ellipse are (±1, 0)c)equation of ellipse is x2 + 2y2 = 4d)the foci of ellipse are (±, 0)Correct answer is option 'A,B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An ellipse intersects the hyperbola 2x2 – 2y2 = 1 orthogonally.The eccentricity of the ellipse is reciprocal of that of the hyperbola. If the axes of the ellipse are along the coordinate axes, then (2009)a)equation of ellipse is x2 + 2y2 = 2b)the foci of ellipse are (±1, 0)c)equation of ellipse is x2 + 2y2 = 4d)the foci of ellipse are (±, 0)Correct answer is option 'A,B'. Can you explain this answer?.
An ellipse intersects the hyperbola 2x2 – 2y2 = 1 orthogonally.The eccentricity of the ellipse is reciprocal of that of the hyperbola. If the axes of the ellipse are along the coordinate axes, then (2009)a)equation of ellipse is x2 + 2y2 = 2b)the foci of ellipse are (±1, 0)c)equation of ellipse is x2 + 2y2 = 4d)the foci of ellipse are (±, 0)Correct answer is option 'A,B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about An ellipse intersects the hyperbola 2x2 – 2y2 = 1 orthogonally.The eccentricity of the ellipse is reciprocal of that of the hyperbola. If the axes of the ellipse are along the coordinate axes, then (2009)a)equation of ellipse is x2 + 2y2 = 2b)the foci of ellipse are (±1, 0)c)equation of ellipse is x2 + 2y2 = 4d)the foci of ellipse are (±, 0)Correct answer is option 'A,B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An ellipse intersects the hyperbola 2x2 – 2y2 = 1 orthogonally.The eccentricity of the ellipse is reciprocal of that of the hyperbola. If the axes of the ellipse are along the coordinate axes, then (2009)a)equation of ellipse is x2 + 2y2 = 2b)the foci of ellipse are (±1, 0)c)equation of ellipse is x2 + 2y2 = 4d)the foci of ellipse are (±, 0)Correct answer is option 'A,B'. Can you explain this answer?.
Solutions for An ellipse intersects the hyperbola 2x2 – 2y2 = 1 orthogonally.The eccentricity of the ellipse is reciprocal of that of the hyperbola. If the axes of the ellipse are along the coordinate axes, then (2009)a)equation of ellipse is x2 + 2y2 = 2b)the foci of ellipse are (±1, 0)c)equation of ellipse is x2 + 2y2 = 4d)the foci of ellipse are (±, 0)Correct answer is option 'A,B'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
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An ellipse intersects the hyperbola 2x2 – 2y2 = 1 orthogonally.The eccentricity of the ellipse is reciprocal of that of the hyperbola. If the axes of the ellipse are along the coordinate axes, then (2009)a)equation of ellipse is x2 + 2y2 = 2b)the foci of ellipse are (±1, 0)c)equation of ellipse is x2 + 2y2 = 4d)the foci of ellipse are (±, 0)Correct answer is option 'A,B'. Can you explain this answer?, a detailed solution for An ellipse intersects the hyperbola 2x2 – 2y2 = 1 orthogonally.The eccentricity of the ellipse is reciprocal of that of the hyperbola. If the axes of the ellipse are along the coordinate axes, then (2009)a)equation of ellipse is x2 + 2y2 = 2b)the foci of ellipse are (±1, 0)c)equation of ellipse is x2 + 2y2 = 4d)the foci of ellipse are (±, 0)Correct answer is option 'A,B'. Can you explain this answer? has been provided alongside types of An ellipse intersects the hyperbola 2x2 – 2y2 = 1 orthogonally.The eccentricity of the ellipse is reciprocal of that of the hyperbola. If the axes of the ellipse are along the coordinate axes, then (2009)a)equation of ellipse is x2 + 2y2 = 2b)the foci of ellipse are (±1, 0)c)equation of ellipse is x2 + 2y2 = 4d)the foci of ellipse are (±, 0)Correct answer is option 'A,B'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice An ellipse intersects the hyperbola 2x2 – 2y2 = 1 orthogonally.The eccentricity of the ellipse is reciprocal of that of the hyperbola. If the axes of the ellipse are along the coordinate axes, then (2009)a)equation of ellipse is x2 + 2y2 = 2b)the foci of ellipse are (±1, 0)c)equation of ellipse is x2 + 2y2 = 4d)the foci of ellipse are (±, 0)Correct answer is option 'A,B'. Can you explain this answer? tests, examples and also practice JEE tests.
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