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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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