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Differential equation of the family of ellipses having foci on y-axis and centre at origin is
- a)xy′′ + x(y′)2 yy′ = 0
- b)xyy′′ + x(y′)2 yy′ = 0
- c)yy′′ + x(y′)2 yy′ = 0
- d)xyy′′ −x(y′)2 +yy′ = 0
Correct answer is option 'B'. Can you explain this answer?
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The differential equation of the family of ellipses having foci on the y-axis and center at the origin can be found as follows:
Let's consider an ellipse with foci (0, c) and (0, -c), where c is a positive constant. The distance between the foci is 2c.
The general equation of an ellipse centered at the origin is given by:
x^2/a^2 + y^2/b^2 = 1,
where a and b are positive constants representing the semi-major and semi-minor axes, respectively.
Since the foci lie on the y-axis, the equation of the ellipse becomes:
x^2/a^2 + (y - c)^2/b^2 = 1.
We know that the distance between the foci is 2c, so we have:
2c = 2b^2/a.
Simplifying this equation, we get:
b^2 = ac.
Now, differentiating both sides of the equation with respect to x, we have:
2b(b') = a'c + ac',
where b' and a' represent the derivatives of b and a with respect to x, respectively.
Since the derivatives of a and b are unknown, we cannot solve for the differential equation in terms of x and y directly. However, we can eliminate a' and b' by using the relationship b^2 = ac, which gives:
2b(b') = 2b(c'/b) + ac'.
Simplifying this equation, we find:
b' = c'/b + ac'/(2b^2).
Rearranging terms, we get:
b'(2b^2) = c' + ac'.
Substituting b^2 = ac, we obtain:
b'(2b^2) = c' + b^2c'.
Finally, dividing both sides by 2b^2, we get the differential equation:
b' = (c' + b^2c')/(2b^2).
Therefore, the differential equation of the family of ellipses having foci on the y-axis and center at the origin is:
b' = (c' + b^2c')/(2b^2).
Let's consider an ellipse with foci (0, c) and (0, -c), where c is a positive constant. The distance between the foci is 2c.
The general equation of an ellipse centered at the origin is given by:
x^2/a^2 + y^2/b^2 = 1,
where a and b are positive constants representing the semi-major and semi-minor axes, respectively.
Since the foci lie on the y-axis, the equation of the ellipse becomes:
x^2/a^2 + (y - c)^2/b^2 = 1.
We know that the distance between the foci is 2c, so we have:
2c = 2b^2/a.
Simplifying this equation, we get:
b^2 = ac.
Now, differentiating both sides of the equation with respect to x, we have:
2b(b') = a'c + ac',
where b' and a' represent the derivatives of b and a with respect to x, respectively.
Since the derivatives of a and b are unknown, we cannot solve for the differential equation in terms of x and y directly. However, we can eliminate a' and b' by using the relationship b^2 = ac, which gives:
2b(b') = 2b(c'/b) + ac'.
Simplifying this equation, we find:
b' = c'/b + ac'/(2b^2).
Rearranging terms, we get:
b'(2b^2) = c' + ac'.
Substituting b^2 = ac, we obtain:
b'(2b^2) = c' + b^2c'.
Finally, dividing both sides by 2b^2, we get the differential equation:
b' = (c' + b^2c')/(2b^2).
Therefore, the differential equation of the family of ellipses having foci on the y-axis and center at the origin is:
b' = (c' + b^2c')/(2b^2).
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Differential equation of the family of ellipses having foci on y-axis and centre at origin isa)xy′′+x(y′)2yy′=0b)xyy′′+x(y′)2yy′=0c)yy′′+x(y′)2yy′=0d)xyy′′−x(y′)2+yy′=0Correct answer is option 'B'. Can you explain this answer?
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Differential equation of the family of ellipses having foci on y-axis and centre at origin isa)xy′′+x(y′)2yy′=0b)xyy′′+x(y′)2yy′=0c)yy′′+x(y′)2yy′=0d)xyy′′−x(y′)2+yy′=0Correct 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 Differential equation of the family of ellipses having foci on y-axis and centre at origin isa)xy′′+x(y′)2yy′=0b)xyy′′+x(y′)2yy′=0c)yy′′+x(y′)2yy′=0d)xyy′′−x(y′)2+yy′=0Correct 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 Differential equation of the family of ellipses having foci on y-axis and centre at origin isa)xy′′+x(y′)2yy′=0b)xyy′′+x(y′)2yy′=0c)yy′′+x(y′)2yy′=0d)xyy′′−x(y′)2+yy′=0Correct answer is option 'B'. Can you explain this answer?.
Differential equation of the family of ellipses having foci on y-axis and centre at origin isa)xy′′+x(y′)2yy′=0b)xyy′′+x(y′)2yy′=0c)yy′′+x(y′)2yy′=0d)xyy′′−x(y′)2+yy′=0Correct 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 Differential equation of the family of ellipses having foci on y-axis and centre at origin isa)xy′′+x(y′)2yy′=0b)xyy′′+x(y′)2yy′=0c)yy′′+x(y′)2yy′=0d)xyy′′−x(y′)2+yy′=0Correct 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 Differential equation of the family of ellipses having foci on y-axis and centre at origin isa)xy′′+x(y′)2yy′=0b)xyy′′+x(y′)2yy′=0c)yy′′+x(y′)2yy′=0d)xyy′′−x(y′)2+yy′=0Correct answer is option 'B'. Can you explain this answer?.
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