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The differential equation for the family of curves x2 + y2 - 2ay = 0, where a is an arbitrary constant is
- a)(x2 - y2)y′ = 2xy
- b)2(x2 + y2)y′ = xy
- c)2(x2 - y2)y′ = xy
- d)(x2 + y2)y′ = 2xy
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
The differential equation for the family of curves x2 + y2 - 2ay = 0, ...
Understanding the Family of Curves
The given family of curves is represented by the equation:
- x² + y² - 2ay = 0, where 'a' is an arbitrary constant.
To find the differential equation corresponding to this family of curves, we need to eliminate the constant 'a'.
Step 1: Rewrite the Equation
We can rearrange the equation as follows:
- y² - 2ay + x² = 0.
This is a quadratic equation in 'y'. We can apply the quadratic formula:
- y = a ± √(a² - x²).
Step 2: Differentiate Implicitly
Now, we differentiate both sides with respect to 'x':
- d/dx(y²) - 2a(d/dx(y)) + d/dx(x²) = 0.
This leads to:
- 2y(dy/dx) - 2a(dy/dx) + 2x = 0.
Rearranging gives us:
- (2y - 2a)(dy/dx) = -2x.
From this, we can express dy/dx:
- dy/dx = -2x / (2y - 2a) = -x / (y - a).
Step 3: Eliminate 'a'
Since a = (y + √(y² - x²))/2, we can substitute this back into our equation to eliminate 'a'.
After simplification, we arrive at the expression:
- (x² - y²)dy/dx = 2xy.
This matches the form of option 'A'.
Conclusion
Thus, the correct differential equation for the family of curves is:
- (x² - y²)dy/dx = 2xy, which corresponds to option 'A'.
The given family of curves is represented by the equation:
- x² + y² - 2ay = 0, where 'a' is an arbitrary constant.
To find the differential equation corresponding to this family of curves, we need to eliminate the constant 'a'.
Step 1: Rewrite the Equation
We can rearrange the equation as follows:
- y² - 2ay + x² = 0.
This is a quadratic equation in 'y'. We can apply the quadratic formula:
- y = a ± √(a² - x²).
Step 2: Differentiate Implicitly
Now, we differentiate both sides with respect to 'x':
- d/dx(y²) - 2a(d/dx(y)) + d/dx(x²) = 0.
This leads to:
- 2y(dy/dx) - 2a(dy/dx) + 2x = 0.
Rearranging gives us:
- (2y - 2a)(dy/dx) = -2x.
From this, we can express dy/dx:
- dy/dx = -2x / (2y - 2a) = -x / (y - a).
Step 3: Eliminate 'a'
Since a = (y + √(y² - x²))/2, we can substitute this back into our equation to eliminate 'a'.
After simplification, we arrive at the expression:
- (x² - y²)dy/dx = 2xy.
This matches the form of option 'A'.
Conclusion
Thus, the correct differential equation for the family of curves is:
- (x² - y²)dy/dx = 2xy, which corresponds to option 'A'.
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The differential equation for the family of curves x2 + y2 - 2ay = 0, where a is an arbitrary constant isa)(x2 - y2)y′ = 2xyb)2(x2 + y2)y′ = xyc)2(x2 - y2)y′ = xyd)(x2 + y2)y′ = 2xyCorrect answer is option 'A'. Can you explain this answer?
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The differential equation for the family of curves x2 + y2 - 2ay = 0, where a is an arbitrary constant isa)(x2 - y2)y′ = 2xyb)2(x2 + y2)y′ = xyc)2(x2 - y2)y′ = xyd)(x2 + y2)y′ = 2xyCorrect answer is option 'A'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The differential equation for the family of curves x2 + y2 - 2ay = 0, where a is an arbitrary constant isa)(x2 - y2)y′ = 2xyb)2(x2 + y2)y′ = xyc)2(x2 - y2)y′ = xyd)(x2 + y2)y′ = 2xyCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The differential equation for the family of curves x2 + y2 - 2ay = 0, where a is an arbitrary constant isa)(x2 - y2)y′ = 2xyb)2(x2 + y2)y′ = xyc)2(x2 - y2)y′ = xyd)(x2 + y2)y′ = 2xyCorrect answer is option 'A'. Can you explain this answer?.
The differential equation for the family of curves x2 + y2 - 2ay = 0, where a is an arbitrary constant isa)(x2 - y2)y′ = 2xyb)2(x2 + y2)y′ = xyc)2(x2 - y2)y′ = xyd)(x2 + y2)y′ = 2xyCorrect answer is option 'A'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The differential equation for the family of curves x2 + y2 - 2ay = 0, where a is an arbitrary constant isa)(x2 - y2)y′ = 2xyb)2(x2 + y2)y′ = xyc)2(x2 - y2)y′ = xyd)(x2 + y2)y′ = 2xyCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The differential equation for the family of curves x2 + y2 - 2ay = 0, where a is an arbitrary constant isa)(x2 - y2)y′ = 2xyb)2(x2 + y2)y′ = xyc)2(x2 - y2)y′ = xyd)(x2 + y2)y′ = 2xyCorrect answer is option 'A'. Can you explain this answer?.
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