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The equation of a line passing through the centre of a rectangular hyperbola is x – y –1 = 0. if one of its asymptote is 3x-4y-6 = 0, then equation of its other asymptote is
- a)4x + 3y + 17 = 0
- b)4x + 3y -17 = 0
- c)4x + 3y -15 = 0
- d)4x + 3y + 15 = 0
Correct answer is option 'B'. Can you explain this answer?
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The equation of a line passing through the centre of a rectangular hyp...
Pt of intersection of x-y-1=0 and 3x-4y-6=0 is (-2,-3) other asymptote will be in the form of 4x + 3y + λ =0 and it should pass through (-2,-3). Thus λ = -17
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The equation of a line passing through the centre of a rectangular hyp...
Given Information:
- Equation of a line passing through the centre of a rectangular hyperbola: x - y - 1 = 0
- One of its asymptotes: 3x - 4y - 6 = 0
Explanation:
The equation of the hyperbola's asymptotes can be written in the form of ax + by + c = 0, where a, b, and c are constants.
Equation of the first asymptote:
Given: 3x - 4y - 6 = 0
This represents one of the asymptotes of the hyperbola.
Using the formula for the equation of the other asymptote:
For a rectangular hyperbola, the equations of the asymptotes are of the form:
(x^2 / a^2) - (y^2 / b^2) = 1
The equations of the asymptotes are:
y = (b/a)x
y = -(b/a)x
Equation of the second asymptote:
Given that the line x - y - 1 = 0 passes through the centre of the hyperbola, we can find the slope of this line, which is 1.
The slope of the first asymptote is 3/4, and the slope of the second asymptote will be the negative reciprocal of the slope of the first asymptote, which is -4/3.
Therefore, the equation of the second asymptote will be of the form 4x + 3y + k = 0, where k is a constant to be determined.
Calculating k:
Substitute a point on the hyperbola into the equation:
(0, -6) satisfies 3x - 4y - 6 = 0
4(0) + 3(-6) + k = 0
-18 + k = 0
k = 18
Therefore, the equation of the second asymptote is 4x + 3y - 18 = 0, which can be simplified to 4x + 3y - 18 = 0.
Thus, the correct answer is option B: 4x + 3y - 18 = 0.
- Equation of a line passing through the centre of a rectangular hyperbola: x - y - 1 = 0
- One of its asymptotes: 3x - 4y - 6 = 0
Explanation:
The equation of the hyperbola's asymptotes can be written in the form of ax + by + c = 0, where a, b, and c are constants.
Equation of the first asymptote:
Given: 3x - 4y - 6 = 0
This represents one of the asymptotes of the hyperbola.
Using the formula for the equation of the other asymptote:
For a rectangular hyperbola, the equations of the asymptotes are of the form:
(x^2 / a^2) - (y^2 / b^2) = 1
The equations of the asymptotes are:
y = (b/a)x
y = -(b/a)x
Equation of the second asymptote:
Given that the line x - y - 1 = 0 passes through the centre of the hyperbola, we can find the slope of this line, which is 1.
The slope of the first asymptote is 3/4, and the slope of the second asymptote will be the negative reciprocal of the slope of the first asymptote, which is -4/3.
Therefore, the equation of the second asymptote will be of the form 4x + 3y + k = 0, where k is a constant to be determined.
Calculating k:
Substitute a point on the hyperbola into the equation:
(0, -6) satisfies 3x - 4y - 6 = 0
4(0) + 3(-6) + k = 0
-18 + k = 0
k = 18
Therefore, the equation of the second asymptote is 4x + 3y - 18 = 0, which can be simplified to 4x + 3y - 18 = 0.
Thus, the correct answer is option B: 4x + 3y - 18 = 0.
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The equation of a line passing through the centre of a rectangular hyperbola is x – y –1 = 0. if one of its asymptote is 3x-4y-6 = 0, then equation of its other asymptoteisa)4x + 3y + 17 = 0b)4x + 3y -17 = 0c)4x + 3y -15 = 0d)4x + 3y + 15 = 0Correct answer is option 'B'. Can you explain this answer?
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The equation of a line passing through the centre of a rectangular hyperbola is x – y –1 = 0. if one of its asymptote is 3x-4y-6 = 0, then equation of its other asymptoteisa)4x + 3y + 17 = 0b)4x + 3y -17 = 0c)4x + 3y -15 = 0d)4x + 3y + 15 = 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 The equation of a line passing through the centre of a rectangular hyperbola is x – y –1 = 0. if one of its asymptote is 3x-4y-6 = 0, then equation of its other asymptoteisa)4x + 3y + 17 = 0b)4x + 3y -17 = 0c)4x + 3y -15 = 0d)4x + 3y + 15 = 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 The equation of a line passing through the centre of a rectangular hyperbola is x – y –1 = 0. if one of its asymptote is 3x-4y-6 = 0, then equation of its other asymptoteisa)4x + 3y + 17 = 0b)4x + 3y -17 = 0c)4x + 3y -15 = 0d)4x + 3y + 15 = 0Correct answer is option 'B'. Can you explain this answer?.
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