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If the eccentricity of the hyperbola x2 – y2 sec2 a = 5 is (√3) times the eccentricity of the ellipse x2 sec2 a + y2 = 25, then a value e of a is
- a)π/6
- b)π/4
- c)π/3
- d)π/2
Correct answer is option 'B'. Can you explain this answer?
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If the eccentricity of the hyperbola x2–y2sec2a = 5 is (√3...
Given Information:
The eccentricity of the hyperbola x^2 - y^2 sec^2a = 5 is √3 times the eccentricity of the ellipse x^2 sec^2a + y^2 = 25.
To Find:
The value of 'a' where the eccentricity of the hyperbola is √3 times the eccentricity of the ellipse.
Solution:
Step 1: Find the eccentricity of the hyperbola
For a hyperbola of the form x^2/a^2 - y^2/b^2 = 1, the eccentricity is given by e = √(1 + b^2/a^2).
In this case, the hyperbola is x^2 - y^2 sec^2a = 5.
Comparing, we get a^2 = 5 and b^2 = 1.
So, the eccentricity of the hyperbola is e_hyperbola = √(1 + 1/5) = √(6/5).
Step 2: Find the eccentricity of the ellipse
For an ellipse of the form x^2/a^2 + y^2/b^2 = 1, the eccentricity is given by e = √(1 - b^2/a^2).
In this case, the ellipse is x^2 sec^2a + y^2 = 25.
Comparing, we get a^2 = 25 and b^2 = sec^2a.
So, the eccentricity of the ellipse is e_ellipse = √(1 - sec^2a/25) = √(24/25).
Step 3: Given Condition
It is given that e_hyperbola = √3 * e_ellipse.
Therefore, √(6/5) = √3 * √(24/25).
Solving this equation, we get sec^2a = 8/5.
Step 4: Find the value of 'a'
From sec^2a = 8/5, we get cos^2a = 5/8.
Taking the square root, we get cos a = √(5/8).
Taking the inverse cosine, we get a = π/4.
Therefore, the value of 'a' for the given conditions is π/4. Hence, option 'B' is the correct answer.
The eccentricity of the hyperbola x^2 - y^2 sec^2a = 5 is √3 times the eccentricity of the ellipse x^2 sec^2a + y^2 = 25.
To Find:
The value of 'a' where the eccentricity of the hyperbola is √3 times the eccentricity of the ellipse.
Solution:
Step 1: Find the eccentricity of the hyperbola
For a hyperbola of the form x^2/a^2 - y^2/b^2 = 1, the eccentricity is given by e = √(1 + b^2/a^2).
In this case, the hyperbola is x^2 - y^2 sec^2a = 5.
Comparing, we get a^2 = 5 and b^2 = 1.
So, the eccentricity of the hyperbola is e_hyperbola = √(1 + 1/5) = √(6/5).
Step 2: Find the eccentricity of the ellipse
For an ellipse of the form x^2/a^2 + y^2/b^2 = 1, the eccentricity is given by e = √(1 - b^2/a^2).
In this case, the ellipse is x^2 sec^2a + y^2 = 25.
Comparing, we get a^2 = 25 and b^2 = sec^2a.
So, the eccentricity of the ellipse is e_ellipse = √(1 - sec^2a/25) = √(24/25).
Step 3: Given Condition
It is given that e_hyperbola = √3 * e_ellipse.
Therefore, √(6/5) = √3 * √(24/25).
Solving this equation, we get sec^2a = 8/5.
Step 4: Find the value of 'a'
From sec^2a = 8/5, we get cos^2a = 5/8.
Taking the square root, we get cos a = √(5/8).
Taking the inverse cosine, we get a = π/4.
Therefore, the value of 'a' for the given conditions is π/4. Hence, option 'B' is the correct answer.
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If the eccentricity of the hyperbola x2–y2sec2a = 5 is (√3...
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If the eccentricity of the hyperbola x2–y2sec2a = 5 is (√3) times the eccentricity of the ellipse x2sec2a + y2= 25, then a value e of a isa)π/6b)π/4c)π/3d)π/2Correct answer is option 'B'. Can you explain this answer?
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If the eccentricity of the hyperbola x2–y2sec2a = 5 is (√3) times the eccentricity of the ellipse x2sec2a + y2= 25, then a value e of a isa)π/6b)π/4c)π/3d)π/2Correct answer is option '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 If the eccentricity of the hyperbola x2–y2sec2a = 5 is (√3) times the eccentricity of the ellipse x2sec2a + y2= 25, then a value e of a isa)π/6b)π/4c)π/3d)π/2Correct answer is option '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 If the eccentricity of the hyperbola x2–y2sec2a = 5 is (√3) times the eccentricity of the ellipse x2sec2a + y2= 25, then a value e of a isa)π/6b)π/4c)π/3d)π/2Correct answer is option 'B'. Can you explain this answer?.
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