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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 coordinates axes, then
- 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 (±√2, 0)
Correct answer is option 'A, B'. Can you explain this answer?
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Most Upvoted Answer
An ellipse intersects the hyperbola 2x2 − 2y2 = 1 orthogonally. The e...
Eccentricity of the hyperbola is 2√2 as it is a rectangular hyperbola so eccentricity ee of the ellipse is 1/√2
Let the equation of the ellipse be
So, the equation of the ellipse is
x2 +2y2 = a2x2 + 2y2 = a2
Let (x1, y1) be a point of intersection of the ellipse and the hyperbola, then
Equations of the tangents at (x1, y1) to the two conics are
Since the two conics intersect orthogonally
and from (1), we get
Hence, the equation of the ellipse is x2 + 2y2 = 2 and its focus is
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An ellipse intersects the hyperbola 2x2 − 2y2 = 1 orthogonally. The e...
Given:
- Equation of hyperbola: 2x^2 - 2y^2 = 1
- The ellipse intersects the hyperbola orthogonally.
- The eccentricity of the ellipse is reciprocal of that of the hyperbola.
- The axes of the ellipse are along the coordinate axes.
To Find:
a) Equation of the ellipse
b) The foci of the ellipse
Solution:
To solve this problem, let's analyze the given information step by step.
1. Orthogonal Intersection:
When an ellipse intersects a hyperbola orthogonally, it means the tangents drawn at the points of intersection are perpendicular to each other. In other words, the product of the slopes of the tangents is equal to -1.
The equation of the hyperbola is 2x^2 - 2y^2 = 1. To find the slopes of the tangents, let's differentiate the equation implicitly.
Differentiating both sides with respect to x:
4x - 4yy' = 0
y' = x/y
Now, let's find the slope of the tangent at any point (x, y) on the hyperbola:
m1 = x/y
To find the slope of the other tangent at the point of intersection, we need to differentiate the equation of the ellipse implicitly.
2. Reciprocal Eccentricity:
The eccentricity of an ellipse is given by the formula:
e = sqrt(1 - (b^2/a^2))
The eccentricity of the hyperbola is the reciprocal of the ellipse's eccentricity. Therefore, we can write:
1/e = sqrt(1 - (a^2/b^2))
Simplifying the equation, we get:
e = b/a
3. Finding the Ellipse Equation:
Since the axes of the ellipse are along the coordinate axes, its standard equation can be written as:
x^2/a^2 + y^2/b^2 = 1
Comparing this equation with the given hyperbola equation, we can see that a^2 = 1/2 and b^2 = 1/2.
Therefore, the equation of the ellipse is:
x^2/(1/2) + y^2/(1/2) = 1
2x^2 + 2y^2 = 2
x^2 + y^2 = 1
So, option A is correct: the equation of the ellipse is x^2 + y^2 = 1.
4. Finding the Foci of the Ellipse:
The foci of an ellipse can be found using the formula:
c = sqrt(a^2 - b^2)
Substituting the values of a and b, we get:
c = sqrt(1/2 - 1/2) = 0
Since the value of c is 0, the foci of the ellipse lie at the origin (0, 0). Therefore, option B is correct: the foci of the ellipse are (±1, 0).
Hence, the correct answers are option A and option B.
- Equation of hyperbola: 2x^2 - 2y^2 = 1
- The ellipse intersects the hyperbola orthogonally.
- The eccentricity of the ellipse is reciprocal of that of the hyperbola.
- The axes of the ellipse are along the coordinate axes.
To Find:
a) Equation of the ellipse
b) The foci of the ellipse
Solution:
To solve this problem, let's analyze the given information step by step.
1. Orthogonal Intersection:
When an ellipse intersects a hyperbola orthogonally, it means the tangents drawn at the points of intersection are perpendicular to each other. In other words, the product of the slopes of the tangents is equal to -1.
The equation of the hyperbola is 2x^2 - 2y^2 = 1. To find the slopes of the tangents, let's differentiate the equation implicitly.
Differentiating both sides with respect to x:
4x - 4yy' = 0
y' = x/y
Now, let's find the slope of the tangent at any point (x, y) on the hyperbola:
m1 = x/y
To find the slope of the other tangent at the point of intersection, we need to differentiate the equation of the ellipse implicitly.
2. Reciprocal Eccentricity:
The eccentricity of an ellipse is given by the formula:
e = sqrt(1 - (b^2/a^2))
The eccentricity of the hyperbola is the reciprocal of the ellipse's eccentricity. Therefore, we can write:
1/e = sqrt(1 - (a^2/b^2))
Simplifying the equation, we get:
e = b/a
3. Finding the Ellipse Equation:
Since the axes of the ellipse are along the coordinate axes, its standard equation can be written as:
x^2/a^2 + y^2/b^2 = 1
Comparing this equation with the given hyperbola equation, we can see that a^2 = 1/2 and b^2 = 1/2.
Therefore, the equation of the ellipse is:
x^2/(1/2) + y^2/(1/2) = 1
2x^2 + 2y^2 = 2
x^2 + y^2 = 1
So, option A is correct: the equation of the ellipse is x^2 + y^2 = 1.
4. Finding the Foci of the Ellipse:
The foci of an ellipse can be found using the formula:
c = sqrt(a^2 - b^2)
Substituting the values of a and b, we get:
c = sqrt(1/2 - 1/2) = 0
Since the value of c is 0, the foci of the ellipse lie at the origin (0, 0). Therefore, option B is correct: the foci of the ellipse are (±1, 0).
Hence, the correct answers are option A and option B.
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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 coordinates axes, thena)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 (±√2, 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 coordinates axes, thena)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 (±√2, 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 coordinates axes, thena)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 (±√2, 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 coordinates axes, thena)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 (±√2, 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 coordinates axes, thena)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 (±√2, 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 coordinates axes, thena)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 (±√2, 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 coordinates axes, thena)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 (±√2, 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 coordinates axes, thena)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 (±√2, 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 coordinates axes, thena)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 (±√2, 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 coordinates axes, thena)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 (±√2, 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 coordinates axes, thena)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 (±√2, 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 coordinates axes, thena)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 (±√2, 0)Correct answer is option 'A, B'. Can you explain this answer? tests, examples and also practice JEE tests.
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