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The equation of the curve for which the square of the ordinate is twice the rectangle contained by the abscissa and the intercept of the normal on x-axis and passing through (2, 1) is
- a)x2 + y2 − x = 0
- b)4x2 + 2y2 − 9y = 0
- c)2x2 + 4y2 − 9x = 0
- d)4x2 + 2y2 − 9x = 0
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
The equation of the curve for which the square of the ordinate is twic...
∵ Equation of normal at (x, y) is
Y − y = dx/dy (X − x)
Put, y = 0
Then, X = x + y dy/dx
Given, y2 = 2x X
⇒ y2 = 2x ( x + y dy/dx)
⇒ dy/dx = (y2 − 2x2)/(2xy) = ((y/x)2 − 2)/ ( 2y/x)
Put y = vx, we get dx/dy = v + x dv/dx
Then, v + x dv/dx = v2−2/2v
On integrating both sides, we get
ln (2 + v2) + ln|x| = ln c
⇒ ln (|x|(2 + v2)) = ln c
⇒ |x| ( 2 + y2/x2) = c
∵ It passes through (2, 1), then 2 (2 + 1 4 ) = c
⇒ c = 9/2
Then, |x| ( 2 + y2/x2) = 9/2
⇒ 2x2 + y2 = 9/2 |x|
⇒ 4x2 + 2y2 = 9|x|
Y − y = dx/dy (X − x)
Put, y = 0
Then, X = x + y dy/dx
Given, y2 = 2x X
⇒ y2 = 2x ( x + y dy/dx)
⇒ dy/dx = (y2 − 2x2)/(2xy) = ((y/x)2 − 2)/ ( 2y/x)
Put y = vx, we get dx/dy = v + x dv/dx
Then, v + x dv/dx = v2−2/2v
On integrating both sides, we get
ln (2 + v2) + ln|x| = ln c
⇒ ln (|x|(2 + v2)) = ln c
⇒ |x| ( 2 + y2/x2) = c
∵ It passes through (2, 1), then 2 (2 + 1 4 ) = c
⇒ c = 9/2
Then, |x| ( 2 + y2/x2) = 9/2
⇒ 2x2 + y2 = 9/2 |x|
⇒ 4x2 + 2y2 = 9|x|
Most Upvoted Answer
The equation of the curve for which the square of the ordinate is twic...
Let the equation of the curve be y=f(x).
The abscissa of the point where the normal intersects the x-axis is also the x-coordinate of the point where the normal intersects the curve. Let this point be (a, 0).
We know that the equation of the normal at any point (x, y) on the curve is given by y - f(x) = f'(x)(x - x). Since the normal passes through (a, 0), we have 0 - f(a) = f'(a)(a - a).
This simplifies to -f(a) = 0, which means f(a) = 0.
Thus, the abscissa of the point where the normal intersects the x-axis is the root of the equation f(x) = 0.
Given that the square of the ordinate is twice the rectangle contained by the abscissa and the intercept of the normal on the x-axis, we have y^2 = 2(x - a)(a - 0).
Since the normal passes through (2, 1), we have 1 = f'(a)(2 - a).
Simplifying, we get f'(a) = 1/(2 - a).
To find the equation of the curve, we need to integrate f'(x) with respect to x.
Integrating f'(x) = 1/(2 - x), we get f(x) = -ln|2 - x| + C, where C is the constant of integration.
Since f(a) = 0, we have -ln|2 - a| + C = 0.
Thus, C = ln|2 - a|.
Therefore, the equation of the curve is f(x) = -ln|2 - x| + ln|2 - a|.
Simplifying, we get f(x) = ln|(2 - a)/(2 - x)|.
Therefore, the equation of the curve is y = ln|(2 - a)/(2 - x)|.
The abscissa of the point where the normal intersects the x-axis is also the x-coordinate of the point where the normal intersects the curve. Let this point be (a, 0).
We know that the equation of the normal at any point (x, y) on the curve is given by y - f(x) = f'(x)(x - x). Since the normal passes through (a, 0), we have 0 - f(a) = f'(a)(a - a).
This simplifies to -f(a) = 0, which means f(a) = 0.
Thus, the abscissa of the point where the normal intersects the x-axis is the root of the equation f(x) = 0.
Given that the square of the ordinate is twice the rectangle contained by the abscissa and the intercept of the normal on the x-axis, we have y^2 = 2(x - a)(a - 0).
Since the normal passes through (2, 1), we have 1 = f'(a)(2 - a).
Simplifying, we get f'(a) = 1/(2 - a).
To find the equation of the curve, we need to integrate f'(x) with respect to x.
Integrating f'(x) = 1/(2 - x), we get f(x) = -ln|2 - x| + C, where C is the constant of integration.
Since f(a) = 0, we have -ln|2 - a| + C = 0.
Thus, C = ln|2 - a|.
Therefore, the equation of the curve is f(x) = -ln|2 - x| + ln|2 - a|.
Simplifying, we get f(x) = ln|(2 - a)/(2 - x)|.
Therefore, the equation of the curve is y = ln|(2 - a)/(2 - x)|.
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Community Answer
The equation of the curve for which the square of the ordinate is twic...
∵ Equation of normal at (x, y) is
Y − y = dx/dy (X − x)
Put, y = 0
Then, X = x + y dy/dx
Given, y2 = 2x X
⇒ y2 = 2x ( x + y dy/dx)
⇒ dy/dx = (y2 − 2x2)/(2xy) = ((y/x)2 − 2)/ ( 2y/x)
Put y = vx, we get dx/dy = v + x dv/dx
Then, v + x dv/dx = v2−2/2v
On integrating both sides, we get
ln (2 + v2) + ln|x| = ln c
⇒ ln (|x|(2 + v2)) = ln c
⇒ |x| ( 2 + y2/x2) = c
∵ It passes through (2, 1), then 2 (2 + 1 4 ) = c
⇒ c = 9/2
Then, |x| ( 2 + y2/x2) = 9/2
⇒ 2x2 + y2 = 9/2 |x|
⇒ 4x2 + 2y2 = 9|x|
Y − y = dx/dy (X − x)
Put, y = 0
Then, X = x + y dy/dx
Given, y2 = 2x X
⇒ y2 = 2x ( x + y dy/dx)
⇒ dy/dx = (y2 − 2x2)/(2xy) = ((y/x)2 − 2)/ ( 2y/x)
Put y = vx, we get dx/dy = v + x dv/dx
Then, v + x dv/dx = v2−2/2v
On integrating both sides, we get
ln (2 + v2) + ln|x| = ln c
⇒ ln (|x|(2 + v2)) = ln c
⇒ |x| ( 2 + y2/x2) = c
∵ It passes through (2, 1), then 2 (2 + 1 4 ) = c
⇒ c = 9/2
Then, |x| ( 2 + y2/x2) = 9/2
⇒ 2x2 + y2 = 9/2 |x|
⇒ 4x2 + 2y2 = 9|x|
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The equation of the curve for which the square of the ordinate is twice the rectangle contained by the abscissa and the intercept of the normal on x-axis and passing through (2, 1) isa)x2 + y2 − x = 0b)4x2 + 2y2 − 9y = 0c)2x2 + 4y2 − 9x = 0d)4x2 + 2y2 − 9x = 0Correct answer is option 'D'. Can you explain this answer?
Question Description
The equation of the curve for which the square of the ordinate is twice the rectangle contained by the abscissa and the intercept of the normal on x-axis and passing through (2, 1) isa)x2 + y2 − x = 0b)4x2 + 2y2 − 9y = 0c)2x2 + 4y2 − 9x = 0d)4x2 + 2y2 − 9x = 0Correct answer is option 'D'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about The equation of the curve for which the square of the ordinate is twice the rectangle contained by the abscissa and the intercept of the normal on x-axis and passing through (2, 1) isa)x2 + y2 − x = 0b)4x2 + 2y2 − 9y = 0c)2x2 + 4y2 − 9x = 0d)4x2 + 2y2 − 9x = 0Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The equation of the curve for which the square of the ordinate is twice the rectangle contained by the abscissa and the intercept of the normal on x-axis and passing through (2, 1) isa)x2 + y2 − x = 0b)4x2 + 2y2 − 9y = 0c)2x2 + 4y2 − 9x = 0d)4x2 + 2y2 − 9x = 0Correct answer is option 'D'. Can you explain this answer?.
The equation of the curve for which the square of the ordinate is twice the rectangle contained by the abscissa and the intercept of the normal on x-axis and passing through (2, 1) isa)x2 + y2 − x = 0b)4x2 + 2y2 − 9y = 0c)2x2 + 4y2 − 9x = 0d)4x2 + 2y2 − 9x = 0Correct answer is option 'D'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about The equation of the curve for which the square of the ordinate is twice the rectangle contained by the abscissa and the intercept of the normal on x-axis and passing through (2, 1) isa)x2 + y2 − x = 0b)4x2 + 2y2 − 9y = 0c)2x2 + 4y2 − 9x = 0d)4x2 + 2y2 − 9x = 0Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The equation of the curve for which the square of the ordinate is twice the rectangle contained by the abscissa and the intercept of the normal on x-axis and passing through (2, 1) isa)x2 + y2 − x = 0b)4x2 + 2y2 − 9y = 0c)2x2 + 4y2 − 9x = 0d)4x2 + 2y2 − 9x = 0Correct answer is option 'D'. Can you explain this answer?.
Solutions for The equation of the curve for which the square of the ordinate is twice the rectangle contained by the abscissa and the intercept of the normal on x-axis and passing through (2, 1) isa)x2 + y2 − x = 0b)4x2 + 2y2 − 9y = 0c)2x2 + 4y2 − 9x = 0d)4x2 + 2y2 − 9x = 0Correct answer is option 'D'. Can you explain this answer? in English & in Hindi are available as part of our courses for IIT JAM.
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The equation of the curve for which the square of the ordinate is twice the rectangle contained by the abscissa and the intercept of the normal on x-axis and passing through (2, 1) isa)x2 + y2 − x = 0b)4x2 + 2y2 − 9y = 0c)2x2 + 4y2 − 9x = 0d)4x2 + 2y2 − 9x = 0Correct answer is option 'D'. Can you explain this answer?, a detailed solution for The equation of the curve for which the square of the ordinate is twice the rectangle contained by the abscissa and the intercept of the normal on x-axis and passing through (2, 1) isa)x2 + y2 − x = 0b)4x2 + 2y2 − 9y = 0c)2x2 + 4y2 − 9x = 0d)4x2 + 2y2 − 9x = 0Correct answer is option 'D'. Can you explain this answer? has been provided alongside types of The equation of the curve for which the square of the ordinate is twice the rectangle contained by the abscissa and the intercept of the normal on x-axis and passing through (2, 1) isa)x2 + y2 − x = 0b)4x2 + 2y2 − 9y = 0c)2x2 + 4y2 − 9x = 0d)4x2 + 2y2 − 9x = 0Correct answer is option 'D'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice The equation of the curve for which the square of the ordinate is twice the rectangle contained by the abscissa and the intercept of the normal on x-axis and passing through (2, 1) isa)x2 + y2 − x = 0b)4x2 + 2y2 − 9y = 0c)2x2 + 4y2 − 9x = 0d)4x2 + 2y2 − 9x = 0Correct answer is option 'D'. Can you explain this answer? tests, examples and also practice IIT JAM tests.
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