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Locus of the point of intersection of the perpendicular tangents of the curve y2 + 4y – 6x – 2 = 0 is
- a)2x – 1 = 0
- b)2x + 3 = 0
- c)2y + 3 = 0
- d)2x + 5 = 0
Correct answer is option 'C'. Can you explain this answer?
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Locus of the point of intersection of the perpendicular tangents of th...
Given parabola is, y2+4y−6x−2=0
⇒ y2+4y+4=6x+6=6(x+1)
⇒ (y+2)2 = 6(x+1)
shifting origin to (−1,−2)
Y2 = 4aX where a = 3/2
We know locus of point of intersection of perpendicular tangent is directrix of the parabola itself
Hence required locus is X=−a ⇒ x+1=−3/2
⇒ 2x+5=0
⇒ y2+4y+4=6x+6=6(x+1)
⇒ (y+2)2 = 6(x+1)
shifting origin to (−1,−2)
Y2 = 4aX where a = 3/2
We know locus of point of intersection of perpendicular tangent is directrix of the parabola itself
Hence required locus is X=−a ⇒ x+1=−3/2
⇒ 2x+5=0
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Locus of the point of intersection of the perpendicular tangents of th...
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Locus of the point of intersection of the perpendicular tangents of th...
Let's start by finding the equation of the curve.
The given equation is y^2 - 4y = 0.
Factoring out y, we get y(y - 4) = 0.
So, the curve consists of two lines: y = 0 and y = 4.
Next, let's find the equations of the perpendicular tangents to the curve.
For the line y = 0, the tangent is horizontal.
For the line y = 4, the tangent is also horizontal.
Since the perpendicular tangents are both horizontal, their point of intersection will be at the same y-coordinate.
Therefore, the locus of the point of intersection of the perpendicular tangents is the horizontal line y = k, where k can be any real number.
The given equation is y^2 - 4y = 0.
Factoring out y, we get y(y - 4) = 0.
So, the curve consists of two lines: y = 0 and y = 4.
Next, let's find the equations of the perpendicular tangents to the curve.
For the line y = 0, the tangent is horizontal.
For the line y = 4, the tangent is also horizontal.
Since the perpendicular tangents are both horizontal, their point of intersection will be at the same y-coordinate.
Therefore, the locus of the point of intersection of the perpendicular tangents is the horizontal line y = k, where k can be any real number.
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Locus of the point of intersection of the perpendicular tangents of the curve y2+ 4y –6x –2 = 0 isa)2x –1 = 0b)2x + 3 = 0c)2y + 3 = 0d)2x + 5 = 0Correct answer is option 'C'. Can you explain this answer?
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Locus of the point of intersection of the perpendicular tangents of the curve y2+ 4y –6x –2 = 0 isa)2x –1 = 0b)2x + 3 = 0c)2y + 3 = 0d)2x + 5 = 0Correct answer is option 'C'. 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 Locus of the point of intersection of the perpendicular tangents of the curve y2+ 4y –6x –2 = 0 isa)2x –1 = 0b)2x + 3 = 0c)2y + 3 = 0d)2x + 5 = 0Correct 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 Locus of the point of intersection of the perpendicular tangents of the curve y2+ 4y –6x –2 = 0 isa)2x –1 = 0b)2x + 3 = 0c)2y + 3 = 0d)2x + 5 = 0Correct answer is option 'C'. Can you explain this answer?.
Locus of the point of intersection of the perpendicular tangents of the curve y2+ 4y –6x –2 = 0 isa)2x –1 = 0b)2x + 3 = 0c)2y + 3 = 0d)2x + 5 = 0Correct answer is option 'C'. 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 Locus of the point of intersection of the perpendicular tangents of the curve y2+ 4y –6x –2 = 0 isa)2x –1 = 0b)2x + 3 = 0c)2y + 3 = 0d)2x + 5 = 0Correct 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 Locus of the point of intersection of the perpendicular tangents of the curve y2+ 4y –6x –2 = 0 isa)2x –1 = 0b)2x + 3 = 0c)2y + 3 = 0d)2x + 5 = 0Correct answer is option 'C'. Can you explain this answer?.
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Locus of the point of intersection of the perpendicular tangents of the curve y2+ 4y –6x –2 = 0 isa)2x –1 = 0b)2x + 3 = 0c)2y + 3 = 0d)2x + 5 = 0Correct answer is option 'C'. Can you explain this answer?, a detailed solution for Locus of the point of intersection of the perpendicular tangents of the curve y2+ 4y –6x –2 = 0 isa)2x –1 = 0b)2x + 3 = 0c)2y + 3 = 0d)2x + 5 = 0Correct answer is option 'C'. Can you explain this answer? has been provided alongside types of Locus of the point of intersection of the perpendicular tangents of the curve y2+ 4y –6x –2 = 0 isa)2x –1 = 0b)2x + 3 = 0c)2y + 3 = 0d)2x + 5 = 0Correct answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
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