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A curve passes through (1, 1) such that the triangle formed by the coordinate axes and the tangent at any point of the curve is in the first quadrant and has its area equal to 2. What is the differential equation?
- a)dy/dx = [(xy + 2) ± √(1 + xy)]/ x2
- b)dy/dx = [(xy – 2) ± √(1 + xy)]/ x2
- c)dy/dx = [(xy – 2) ± √(1 – xy)]/ x2
- d)dy/dx = [(xy + 2) ± √(1 – xy)]/ x2
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
A curve passes through (1, 1) such that the triangle formed by the coo...
The equation of tangent to the curve y = f(x), at point (x, y), is
Y – y = dy/dx * (X – x) …..(1)
Where it meets the x axis, Y = 0 and X = (x – y/(dy/dx))
Where it meets the y axis, X = 0 and Y = (y – x/(dy/dx))
Also, the area of the triangle formed by (1) with the coordinate axes is 2, so that,
(x – y/(dy/dx))* (y – x/(dy/dx)) = 4
Or, (y – x/(dy/dx))2 – 4dy/dx = 0
Or, x2(dy/dx)2 – 2(xy – 2)dy/dx + y2 = 0
Solving for dy/dx we get,
dy/dx = [(xy – 2) ± √(1 – xy)]/ x2
Y – y = dy/dx * (X – x) …..(1)
Where it meets the x axis, Y = 0 and X = (x – y/(dy/dx))
Where it meets the y axis, X = 0 and Y = (y – x/(dy/dx))
Also, the area of the triangle formed by (1) with the coordinate axes is 2, so that,
(x – y/(dy/dx))* (y – x/(dy/dx)) = 4
Or, (y – x/(dy/dx))2 – 4dy/dx = 0
Or, x2(dy/dx)2 – 2(xy – 2)dy/dx + y2 = 0
Solving for dy/dx we get,
dy/dx = [(xy – 2) ± √(1 – xy)]/ x2
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Community Answer
A curve passes through (1, 1) such that the triangle formed by the coo...
We can start by finding the equation of the tangent line at any point (x, y) on the curve.
Let's assume the equation of the curve is y = f(x). The slope of the tangent line at any point (x, y) on the curve is given by dy/dx.
We are given that the triangle formed by the coordinate axes and the tangent line has an area of 2. The base of this triangle is x, and the height is y. Therefore, the area of the triangle is (1/2) * x * y.
We can now set up the equation for the area of the triangle:
(1/2) * x * y = 2
Simplifying, we get:
x * y = 4
Now, let's find the derivative of both sides with respect to x:
d/dx (x * y) = d/dx (4)
Using the product rule, we have:
y + x * dy/dx = 0
Simplifying, we get:
dy/dx = -y/x
Therefore, the differential equation for the curve is:
dy/dx = -y/x
Let's assume the equation of the curve is y = f(x). The slope of the tangent line at any point (x, y) on the curve is given by dy/dx.
We are given that the triangle formed by the coordinate axes and the tangent line has an area of 2. The base of this triangle is x, and the height is y. Therefore, the area of the triangle is (1/2) * x * y.
We can now set up the equation for the area of the triangle:
(1/2) * x * y = 2
Simplifying, we get:
x * y = 4
Now, let's find the derivative of both sides with respect to x:
d/dx (x * y) = d/dx (4)
Using the product rule, we have:
y + x * dy/dx = 0
Simplifying, we get:
dy/dx = -y/x
Therefore, the differential equation for the curve is:
dy/dx = -y/x
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A curve passes through (1, 1) such that the triangle formed by the coordinate axes and the tangent at any point of the curve is in the first quadrant and has its area equal to 2. What is the differential equation?a)dy/dx = [(xy + 2) ± √(1 + xy)]/ x2b)dy/dx = [(xy – 2) ± √(1 + xy)]/ x2c)dy/dx = [(xy – 2) ± √(1 – xy)]/ x2d)dy/dx = [(xy + 2) ± √(1 – xy)]/ x2Correct answer is option 'C'. Can you explain this answer?
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A curve passes through (1, 1) such that the triangle formed by the coordinate axes and the tangent at any point of the curve is in the first quadrant and has its area equal to 2. What is the differential equation?a)dy/dx = [(xy + 2) ± √(1 + xy)]/ x2b)dy/dx = [(xy – 2) ± √(1 + xy)]/ x2c)dy/dx = [(xy – 2) ± √(1 – xy)]/ x2d)dy/dx = [(xy + 2) ± √(1 – xy)]/ x2Correct answer is option 'C'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about A curve passes through (1, 1) such that the triangle formed by the coordinate axes and the tangent at any point of the curve is in the first quadrant and has its area equal to 2. What is the differential equation?a)dy/dx = [(xy + 2) ± √(1 + xy)]/ x2b)dy/dx = [(xy – 2) ± √(1 + xy)]/ x2c)dy/dx = [(xy – 2) ± √(1 – xy)]/ x2d)dy/dx = [(xy + 2) ± √(1 – xy)]/ x2Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A curve passes through (1, 1) such that the triangle formed by the coordinate axes and the tangent at any point of the curve is in the first quadrant and has its area equal to 2. What is the differential equation?a)dy/dx = [(xy + 2) ± √(1 + xy)]/ x2b)dy/dx = [(xy – 2) ± √(1 + xy)]/ x2c)dy/dx = [(xy – 2) ± √(1 – xy)]/ x2d)dy/dx = [(xy + 2) ± √(1 – xy)]/ x2Correct answer is option 'C'. Can you explain this answer?.
A curve passes through (1, 1) such that the triangle formed by the coordinate axes and the tangent at any point of the curve is in the first quadrant and has its area equal to 2. What is the differential equation?a)dy/dx = [(xy + 2) ± √(1 + xy)]/ x2b)dy/dx = [(xy – 2) ± √(1 + xy)]/ x2c)dy/dx = [(xy – 2) ± √(1 – xy)]/ x2d)dy/dx = [(xy + 2) ± √(1 – xy)]/ x2Correct answer is option 'C'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about A curve passes through (1, 1) such that the triangle formed by the coordinate axes and the tangent at any point of the curve is in the first quadrant and has its area equal to 2. What is the differential equation?a)dy/dx = [(xy + 2) ± √(1 + xy)]/ x2b)dy/dx = [(xy – 2) ± √(1 + xy)]/ x2c)dy/dx = [(xy – 2) ± √(1 – xy)]/ x2d)dy/dx = [(xy + 2) ± √(1 – xy)]/ x2Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A curve passes through (1, 1) such that the triangle formed by the coordinate axes and the tangent at any point of the curve is in the first quadrant and has its area equal to 2. What is the differential equation?a)dy/dx = [(xy + 2) ± √(1 + xy)]/ x2b)dy/dx = [(xy – 2) ± √(1 + xy)]/ x2c)dy/dx = [(xy – 2) ± √(1 – xy)]/ x2d)dy/dx = [(xy + 2) ± √(1 – xy)]/ x2Correct answer is option 'C'. Can you explain this answer?.
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