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The straight line joining any point P on the parabola y2 = 4ax to the vertex and perpendicular from the focus to the tangent at P, intersect at R, then the equation of the locus of R is
- a)x2 + 2y2 – ax = 0
- b)2x2 + y2 – 2ax = 0
- c)2x2 + 2y2 – ay = 0
- d)2x2 + y2 – 2ay = 0
Correct answer is option 'B'. Can you explain this answer?
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The straight line joining any point P on the parabola y2= 4ax to the v...
Given the equation of parabola is
y2 = 4ax
Let P(at2, 2at) be any point on the parabola.
Equation of tangent at point P is ty = x + at2 where slope of the tangent is 1/t.
Equation of line perpendicular to the tangent passes through (a,0) is given as
∴ y−0 = −t(x−a)
or y = t(a−x) .....(i)
Equation of OP is given by
y−0 = 2/t(x−0) = 0
⇒ y = 2x/t .....(ii)
Eliminating 't' from equations (i) and (ii), we get
y2 = 2x(a−x)
or 2x2 + y2 − 2ax = 0
y2 = 4ax
Let P(at2, 2at) be any point on the parabola.
Equation of tangent at point P is ty = x + at2 where slope of the tangent is 1/t.
Equation of line perpendicular to the tangent passes through (a,0) is given as
∴ y−0 = −t(x−a)
or y = t(a−x) .....(i)
Equation of OP is given by
y−0 = 2/t(x−0) = 0
⇒ y = 2x/t .....(ii)
Eliminating 't' from equations (i) and (ii), we get
y2 = 2x(a−x)
or 2x2 + y2 − 2ax = 0
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The straight line joining any point P on the parabola y2= 4ax to the v...
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The straight line joining any point P on the parabola y2= 4ax to the v...
To find the equation of the locus of R, let's first find the coordinates of R in terms of the parameter t.
Let P be the point (at^2, 2at) on the parabola y^2 = 4ax, where a is a constant.
The equation of the tangent at P is given by y - 2at = -1/(2a)(x - at^2).
The slope of the tangent is -1/(2a), so the slope of the line PR is 2a.
Since the line PR is perpendicular to the line joining the focus to the tangent at P, the slope of the line joining the focus (a, 0) to the tangent at P is -1/(2a).
The equation of the line joining the focus (a, 0) to the tangent at P is y = -1/(2a)(x - a).
The equation of the line PR is y - 2at = 2a(x - at^2).
To find the coordinates of R, we need to solve the system of equations formed by the parabola and the line PR:
y^2 = 4ax
y - 2at = 2a(x - at^2)
Substituting y = 2ax/t into the equation y - 2at = 2a(x - at^2), we get:
2ax/t - 2at = 2a(x - at^2)
2ax - 2at^2 = 2a(x - at^2)
2ax - 2at^2 = 2ax - 2a^2t^2
-2at^2 + 2a^2t^2 = 0
-2at^2 + 2a^2t^2 = 0
-2at^2 + 2a^2t^2 = 0
2a^2t^2 - 2at^2 = 0
2at^2(a - 1) = 0
Since a is a nonzero constant, we have t^2(a - 1) = 0.
This gives us two solutions: t = 0 and t = sqrt(a - 1)/sqrt(a).
Substituting t = 0 into the equation y = 2ax/t, we get y = 0.
So, when t = 0, the point R is (0, 0).
Substituting t = sqrt(a - 1)/sqrt(a) into the equation y = 2ax/t, we get y = 2a(sqrt(a - 1)/sqrt(a)).
So, when t = sqrt(a - 1)/sqrt(a), the point R is (sqrt(a - 1), 2a(sqrt(a - 1)/sqrt(a))).
Therefore, the coordinates of R can be written as R = (sqrt(a - 1), 2a(sqrt(a - 1)/sqrt(a))).
Since the locus of R is the set of all possible points R, the equation of the locus is given by:
x = (sqrt(a - 1))^2
y = (2a(sqrt(a - 1)/sqrt(a)))^2
Simplifying, we get:
x = a - 1
y = 4a(a - 1)/a
Let P be the point (at^2, 2at) on the parabola y^2 = 4ax, where a is a constant.
The equation of the tangent at P is given by y - 2at = -1/(2a)(x - at^2).
The slope of the tangent is -1/(2a), so the slope of the line PR is 2a.
Since the line PR is perpendicular to the line joining the focus to the tangent at P, the slope of the line joining the focus (a, 0) to the tangent at P is -1/(2a).
The equation of the line joining the focus (a, 0) to the tangent at P is y = -1/(2a)(x - a).
The equation of the line PR is y - 2at = 2a(x - at^2).
To find the coordinates of R, we need to solve the system of equations formed by the parabola and the line PR:
y^2 = 4ax
y - 2at = 2a(x - at^2)
Substituting y = 2ax/t into the equation y - 2at = 2a(x - at^2), we get:
2ax/t - 2at = 2a(x - at^2)
2ax - 2at^2 = 2a(x - at^2)
2ax - 2at^2 = 2ax - 2a^2t^2
-2at^2 + 2a^2t^2 = 0
-2at^2 + 2a^2t^2 = 0
-2at^2 + 2a^2t^2 = 0
2a^2t^2 - 2at^2 = 0
2at^2(a - 1) = 0
Since a is a nonzero constant, we have t^2(a - 1) = 0.
This gives us two solutions: t = 0 and t = sqrt(a - 1)/sqrt(a).
Substituting t = 0 into the equation y = 2ax/t, we get y = 0.
So, when t = 0, the point R is (0, 0).
Substituting t = sqrt(a - 1)/sqrt(a) into the equation y = 2ax/t, we get y = 2a(sqrt(a - 1)/sqrt(a)).
So, when t = sqrt(a - 1)/sqrt(a), the point R is (sqrt(a - 1), 2a(sqrt(a - 1)/sqrt(a))).
Therefore, the coordinates of R can be written as R = (sqrt(a - 1), 2a(sqrt(a - 1)/sqrt(a))).
Since the locus of R is the set of all possible points R, the equation of the locus is given by:
x = (sqrt(a - 1))^2
y = (2a(sqrt(a - 1)/sqrt(a)))^2
Simplifying, we get:
x = a - 1
y = 4a(a - 1)/a
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The straight line joining any point P on the parabola y2= 4ax to the vertex and perpendicular from the focus to the tangent at P, intersect at R, then the equation of the locus of R isa)x2+ 2y2–ax = 0b)2x2+ y2–2ax = 0c)2x2+ 2y2–ay = 0d)2x2+ y2–2ay = 0Correct answer is option 'B'. Can you explain this answer?
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The straight line joining any point P on the parabola y2= 4ax to the vertex and perpendicular from the focus to the tangent at P, intersect at R, then the equation of the locus of R isa)x2+ 2y2–ax = 0b)2x2+ y2–2ax = 0c)2x2+ 2y2–ay = 0d)2x2+ y2–2ay = 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 straight line joining any point P on the parabola y2= 4ax to the vertex and perpendicular from the focus to the tangent at P, intersect at R, then the equation of the locus of R isa)x2+ 2y2–ax = 0b)2x2+ y2–2ax = 0c)2x2+ 2y2–ay = 0d)2x2+ y2–2ay = 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 straight line joining any point P on the parabola y2= 4ax to the vertex and perpendicular from the focus to the tangent at P, intersect at R, then the equation of the locus of R isa)x2+ 2y2–ax = 0b)2x2+ y2–2ax = 0c)2x2+ 2y2–ay = 0d)2x2+ y2–2ay = 0Correct answer is option 'B'. Can you explain this answer?.
The straight line joining any point P on the parabola y2= 4ax to the vertex and perpendicular from the focus to the tangent at P, intersect at R, then the equation of the locus of R isa)x2+ 2y2–ax = 0b)2x2+ y2–2ax = 0c)2x2+ 2y2–ay = 0d)2x2+ y2–2ay = 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 straight line joining any point P on the parabola y2= 4ax to the vertex and perpendicular from the focus to the tangent at P, intersect at R, then the equation of the locus of R isa)x2+ 2y2–ax = 0b)2x2+ y2–2ax = 0c)2x2+ 2y2–ay = 0d)2x2+ y2–2ay = 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 straight line joining any point P on the parabola y2= 4ax to the vertex and perpendicular from the focus to the tangent at P, intersect at R, then the equation of the locus of R isa)x2+ 2y2–ax = 0b)2x2+ y2–2ax = 0c)2x2+ 2y2–ay = 0d)2x2+ y2–2ay = 0Correct answer is option 'B'. Can you explain this answer?.
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