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At any point (x, y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (– 4, –3). Find the equation of the curve given that it passes through (–2, 1).a)(x+3)2=y+4b)(x+5)2 =2y+3c)(x+4)2 =y+3d)(x+5)2 =2y+3Correct answer is option 'C'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared
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the JEE exam syllabus. Information about At any point (x, y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (– 4, –3). Find the equation of the curve given that it passes through (–2, 1).a)(x+3)2=y+4b)(x+5)2 =2y+3c)(x+4)2 =y+3d)(x+5)2 =2y+3Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam.
Find important definitions, questions, meanings, examples, exercises and tests below for At any point (x, y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (– 4, –3). Find the equation of the curve given that it passes through (–2, 1).a)(x+3)2=y+4b)(x+5)2 =2y+3c)(x+4)2 =y+3d)(x+5)2 =2y+3Correct answer is option 'C'. Can you explain this answer?.
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Here you can find the meaning of At any point (x, y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (– 4, –3). Find the equation of the curve given that it passes through (–2, 1).a)(x+3)2=y+4b)(x+5)2 =2y+3c)(x+4)2 =y+3d)(x+5)2 =2y+3Correct answer is option 'C'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
At any point (x, y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (– 4, –3). Find the equation of the curve given that it passes through (–2, 1).a)(x+3)2=y+4b)(x+5)2 =2y+3c)(x+4)2 =y+3d)(x+5)2 =2y+3Correct answer is option 'C'. Can you explain this answer?, a detailed solution for At any point (x, y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (– 4, –3). Find the equation of the curve given that it passes through (–2, 1).a)(x+3)2=y+4b)(x+5)2 =2y+3c)(x+4)2 =y+3d)(x+5)2 =2y+3Correct answer is option 'C'. Can you explain this answer? has been provided alongside types of At any point (x, y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (– 4, –3). Find the equation of the curve given that it passes through (–2, 1).a)(x+3)2=y+4b)(x+5)2 =2y+3c)(x+4)2 =y+3d)(x+5)2 =2y+3Correct answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice At any point (x, y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (– 4, –3). Find the equation of the curve given that it passes through (–2, 1).a)(x+3)2=y+4b)(x+5)2 =2y+3c)(x+4)2 =y+3d)(x+5)2 =2y+3Correct answer is option 'C'. Can you explain this answer? tests, examples and also practice JEE tests.