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The area enclosed between the curves y = ax2 and x = ay2(a > 0) and is 1 square unit, then the value of a is:
- a)1/3
- b)1
- c)1/2
- d)1/√3
Correct answer is option 'D'. Can you explain this answer?
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The area enclosed between the curvesy = ax2 and ...
We have, the equation of the curves y = ax2, x = ay2.
Points of intersection of the two curves are O(0, 0) and A(1/a, 1/a).
Given, Area bounded between two curve = 1 square unit
Most Upvoted Answer
The area enclosed between the curvesy = ax2 and ...
Given Information:
The area enclosed between the curves \(y = ax^2\) and \(x = ay^2\) (where \(a > 0\)) is 1 square unit.
Solution:
Step 1: Find the points of intersection
To find the points of intersection, set the two equations equal to each other:
\(ax^2 = ay^2\)
\(x^2 = y^2\)
\(x = y\) or \(x = -y\)
Step 2: Find the limits of integration
The limits of integration will be the x-values where the curves intersect. Since the curves are symmetric about the origin, the limits of integration will be from -x to x.
Step 3: Set up the integral to find the area
The area between the two curves can be found by integrating the difference of the two equations:
\(\int_{-x}^{x} (ay^2 - ax^2) dy = 1\)
Step 4: Solve the integral
\(\int_{-x}^{x} (ay^2 - ax^2) dy = [ay^3/3 - ax^2y]_{-x}^{x} = 1\)
Solving for the integral gives:
\(\frac{2ax^3}{3} = 1\)
Step 5: Find the value of a
Given that the area enclosed is 1 square unit, solve for a:
\(\frac{2a}{3} = 1\)
\(a = \frac{3}{2} = \frac{1}{\sqrt{3}}\)
Therefore, the value of \(a\) is \(\frac{1}{\sqrt{3}}\) or option 'D'.
The area enclosed between the curves \(y = ax^2\) and \(x = ay^2\) (where \(a > 0\)) is 1 square unit.
Solution:
Step 1: Find the points of intersection
To find the points of intersection, set the two equations equal to each other:
\(ax^2 = ay^2\)
\(x^2 = y^2\)
\(x = y\) or \(x = -y\)
Step 2: Find the limits of integration
The limits of integration will be the x-values where the curves intersect. Since the curves are symmetric about the origin, the limits of integration will be from -x to x.
Step 3: Set up the integral to find the area
The area between the two curves can be found by integrating the difference of the two equations:
\(\int_{-x}^{x} (ay^2 - ax^2) dy = 1\)
Step 4: Solve the integral
\(\int_{-x}^{x} (ay^2 - ax^2) dy = [ay^3/3 - ax^2y]_{-x}^{x} = 1\)
Solving for the integral gives:
\(\frac{2ax^3}{3} = 1\)
Step 5: Find the value of a
Given that the area enclosed is 1 square unit, solve for a:
\(\frac{2a}{3} = 1\)
\(a = \frac{3}{2} = \frac{1}{\sqrt{3}}\)
Therefore, the value of \(a\) is \(\frac{1}{\sqrt{3}}\) or option 'D'.
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The area enclosed between the curvesy = ax2 and x = ay2(a > 0) and is1 square unit, then the value ofa is:a)1/3b)1c)1/2d)1/√3Correct answer is option 'D'. Can you explain this answer?
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