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A curve y = f (x) passes through (1, 1) and at P(x, y), tangent cuts the x–axis and y–axis at A and B respectively such that BP : AP = 3 : 1, then
- a)equation of curve is xy' – 3y = 0
- b)normal at (1, 1) is x + 3y = 4
- c)curve passes through (2, 1/8)
- d)equation of curve is xy' + 3y = 0
Correct answer is option 'C,D'. Can you explain this answer?
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A curve y = f (x) passes through (1, 1) and at P(x, y), tangent cuts t...
Tangent to the curve y = f (x) at (x, y) is
∴ cur ve is x3y = 1 , which also passes through
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A curve y = f (x) passes through (1, 1) and at P(x, y), tangent cuts t...
-axis at Q. Find the equation of the curve if the area of triangle OPQ is 6 square units.
To find the equation of the curve, we need to determine the function f(x).
Let's consider a point P(x, y) on the curve. The slope of the tangent line at P can be represented as dy/dx.
Since the tangent line passes through (1, 1), we can write the equation of the tangent line as:
y - 1 = dy/dx(x - 1)
Since the tangent line cuts the x-axis at Q, the y-coordinate of Q is 0. Plugging this into the equation of the tangent line, we get:
0 - 1 = dy/dx(x - 1)
-1 = dy/dx(x - 1)
dy/dx = -1/(x - 1)
Now, let's find the equation of the curve by integrating dy/dx with respect to x.
∫dy/dx dx = ∫(-1/(x - 1))dx
∫dy = -∫(1/(x - 1))dx
y = -ln|x - 1| + C
To find the constant C, we can use the fact that the curve passes through (1, 1). Plugging these values into the equation, we get:
1 = -ln|1 - 1| + C
1 = C
Therefore, the equation of the curve is:
y = -ln|x - 1| + 1
To find the equation of the curve, we need to determine the function f(x).
Let's consider a point P(x, y) on the curve. The slope of the tangent line at P can be represented as dy/dx.
Since the tangent line passes through (1, 1), we can write the equation of the tangent line as:
y - 1 = dy/dx(x - 1)
Since the tangent line cuts the x-axis at Q, the y-coordinate of Q is 0. Plugging this into the equation of the tangent line, we get:
0 - 1 = dy/dx(x - 1)
-1 = dy/dx(x - 1)
dy/dx = -1/(x - 1)
Now, let's find the equation of the curve by integrating dy/dx with respect to x.
∫dy/dx dx = ∫(-1/(x - 1))dx
∫dy = -∫(1/(x - 1))dx
y = -ln|x - 1| + C
To find the constant C, we can use the fact that the curve passes through (1, 1). Plugging these values into the equation, we get:
1 = -ln|1 - 1| + C
1 = C
Therefore, the equation of the curve is:
y = -ln|x - 1| + 1
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A curve y = f (x) passes through (1, 1) and at P(x, y), tangent cuts t...
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A curve y = f (x) passes through (1, 1) and at P(x, y), tangent cuts the x–axis and y–axis at A and B respectively such that BP : AP = 3 : 1, thena)equation of curve is xy' – 3y = 0b)normal at (1, 1) is x + 3y = 4c)curve passes through (2, 1/8)d)equation of curve is xy' + 3y = 0Correct answer is option 'C,D'. Can you explain this answer?
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A curve y = f (x) passes through (1, 1) and at P(x, y), tangent cuts the x–axis and y–axis at A and B respectively such that BP : AP = 3 : 1, thena)equation of curve is xy' – 3y = 0b)normal at (1, 1) is x + 3y = 4c)curve passes through (2, 1/8)d)equation of curve is xy' + 3y = 0Correct answer is option 'C,D'. 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 A curve y = f (x) passes through (1, 1) and at P(x, y), tangent cuts the x–axis and y–axis at A and B respectively such that BP : AP = 3 : 1, thena)equation of curve is xy' – 3y = 0b)normal at (1, 1) is x + 3y = 4c)curve passes through (2, 1/8)d)equation of curve is xy' + 3y = 0Correct answer is option 'C,D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A curve y = f (x) passes through (1, 1) and at P(x, y), tangent cuts the x–axis and y–axis at A and B respectively such that BP : AP = 3 : 1, thena)equation of curve is xy' – 3y = 0b)normal at (1, 1) is x + 3y = 4c)curve passes through (2, 1/8)d)equation of curve is xy' + 3y = 0Correct answer is option 'C,D'. Can you explain this answer?.
A curve y = f (x) passes through (1, 1) and at P(x, y), tangent cuts the x–axis and y–axis at A and B respectively such that BP : AP = 3 : 1, thena)equation of curve is xy' – 3y = 0b)normal at (1, 1) is x + 3y = 4c)curve passes through (2, 1/8)d)equation of curve is xy' + 3y = 0Correct answer is option 'C,D'. 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 A curve y = f (x) passes through (1, 1) and at P(x, y), tangent cuts the x–axis and y–axis at A and B respectively such that BP : AP = 3 : 1, thena)equation of curve is xy' – 3y = 0b)normal at (1, 1) is x + 3y = 4c)curve passes through (2, 1/8)d)equation of curve is xy' + 3y = 0Correct answer is option 'C,D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A curve y = f (x) passes through (1, 1) and at P(x, y), tangent cuts the x–axis and y–axis at A and B respectively such that BP : AP = 3 : 1, thena)equation of curve is xy' – 3y = 0b)normal at (1, 1) is x + 3y = 4c)curve passes through (2, 1/8)d)equation of curve is xy' + 3y = 0Correct answer is option 'C,D'. Can you explain this answer?.
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A curve y = f (x) passes through (1, 1) and at P(x, y), tangent cuts the x–axis and y–axis at A and B respectively such that BP : AP = 3 : 1, thena)equation of curve is xy' – 3y = 0b)normal at (1, 1) is x + 3y = 4c)curve passes through (2, 1/8)d)equation of curve is xy' + 3y = 0Correct answer is option 'C,D'. Can you explain this answer?, a detailed solution for A curve y = f (x) passes through (1, 1) and at P(x, y), tangent cuts the x–axis and y–axis at A and B respectively such that BP : AP = 3 : 1, thena)equation of curve is xy' – 3y = 0b)normal at (1, 1) is x + 3y = 4c)curve passes through (2, 1/8)d)equation of curve is xy' + 3y = 0Correct answer is option 'C,D'. Can you explain this answer? has been provided alongside types of A curve y = f (x) passes through (1, 1) and at P(x, y), tangent cuts the x–axis and y–axis at A and B respectively such that BP : AP = 3 : 1, thena)equation of curve is xy' – 3y = 0b)normal at (1, 1) is x + 3y = 4c)curve passes through (2, 1/8)d)equation of curve is xy' + 3y = 0Correct answer is option 'C,D'. Can you explain this answer? theory, EduRev gives you an
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