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The tangent PT and the normal PN to the parabola y2 = 4ax at a point P on it meet its axis at points T and N, respectively. The locus of the centroid of the triangle PTN is a parabola whose
- a)vertex is (2a/3 , 0)
- b)directrix is x = 0
- c)latus rectum is 2a/3
- d)focus is (a, 0)
Correct answer is option 'A,D'. Can you explain this answer?
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The tangent PT and the normal PN to the parabola y2 = 4ax at a point ...
The given parabola equation is y^2 = 4ax, which represents a parabola with its vertex at the origin. Let's solve this problem step by step:
1. Finding the coordinates of point P:
We know that the tangent to a parabola at a point is perpendicular to the line joining the point to the focus of the parabola. In this case, the focus of the parabola is (a, 0). So, the slope of the line joining P to the focus is given by m = -1/(4a).
The equation of the tangent at point P(x, y) can be written as y = mx + a/m. Substituting the value of m, we get y = (-1/(4a))x + a/(-1/(4a)) = (-1/(4a))x - 4a^2.
Now, substituting y^2 = 4ax in the equation of the tangent, we get (-1/(4a))x - 4a^2 = 4ax. Simplifying this equation, we get x = (2a/3).
Therefore, the coordinates of point P are (2a/3, -4a^2).
2. Finding the coordinates of points T and N:
The tangent PT intersects the x-axis at point T. Since the x-coordinate of point P is (2a/3), the x-coordinate of point T is also (2a/3).
The normal PN intersects the x-axis at point N. Since the equation of the normal is y = 4ax, substituting y = 0 in this equation, we get x = 0. Therefore, the x-coordinate of point N is 0.
Therefore, the coordinates of points T and N are (2a/3, 0) and (0, 0) respectively.
3. Finding the centroid of triangle PTN:
The centroid of a triangle is the point of intersection of its medians. In this case, the medians are the line joining the midpoint of PT to the midpoint of TN, the line joining the midpoint of PT to the midpoint of PN, and the line joining the midpoint of TN to the midpoint of PN.
The coordinates of the midpoint of PT are ((2a/3 + 2a/3)/2, (-4a^2 + 0)/2) = (4a/3, -2a^2).
The coordinates of the midpoint of TN are ((2a/3 + 0)/2, (0 + 0)/2) = (a/3, 0).
The coordinates of the midpoint of PN are ((2a/3 + 0)/2, (-4a^2 + 0)/2) = (a/3, -2a^2).
Therefore, the centroid of triangle PTN is ((4a/3 + a/3 + a/3)/3, (-2a^2 + 0 + (-2a^2))/3) = (2a/3, -4a^2/3).
4. Finding the locus of the centroid:
The locus of points that are equidistant from the focus and the directrix of a parabola is the parabola itself. Since the focus of the given parabola is (a, 0) and the directrix is x
1. Finding the coordinates of point P:
We know that the tangent to a parabola at a point is perpendicular to the line joining the point to the focus of the parabola. In this case, the focus of the parabola is (a, 0). So, the slope of the line joining P to the focus is given by m = -1/(4a).
The equation of the tangent at point P(x, y) can be written as y = mx + a/m. Substituting the value of m, we get y = (-1/(4a))x + a/(-1/(4a)) = (-1/(4a))x - 4a^2.
Now, substituting y^2 = 4ax in the equation of the tangent, we get (-1/(4a))x - 4a^2 = 4ax. Simplifying this equation, we get x = (2a/3).
Therefore, the coordinates of point P are (2a/3, -4a^2).
2. Finding the coordinates of points T and N:
The tangent PT intersects the x-axis at point T. Since the x-coordinate of point P is (2a/3), the x-coordinate of point T is also (2a/3).
The normal PN intersects the x-axis at point N. Since the equation of the normal is y = 4ax, substituting y = 0 in this equation, we get x = 0. Therefore, the x-coordinate of point N is 0.
Therefore, the coordinates of points T and N are (2a/3, 0) and (0, 0) respectively.
3. Finding the centroid of triangle PTN:
The centroid of a triangle is the point of intersection of its medians. In this case, the medians are the line joining the midpoint of PT to the midpoint of TN, the line joining the midpoint of PT to the midpoint of PN, and the line joining the midpoint of TN to the midpoint of PN.
The coordinates of the midpoint of PT are ((2a/3 + 2a/3)/2, (-4a^2 + 0)/2) = (4a/3, -2a^2).
The coordinates of the midpoint of TN are ((2a/3 + 0)/2, (0 + 0)/2) = (a/3, 0).
The coordinates of the midpoint of PN are ((2a/3 + 0)/2, (-4a^2 + 0)/2) = (a/3, -2a^2).
Therefore, the centroid of triangle PTN is ((4a/3 + a/3 + a/3)/3, (-2a^2 + 0 + (-2a^2))/3) = (2a/3, -4a^2/3).
4. Finding the locus of the centroid:
The locus of points that are equidistant from the focus and the directrix of a parabola is the parabola itself. Since the focus of the given parabola is (a, 0) and the directrix is x
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The tangent PT and the normal PN to the parabola y2 = 4ax at a point ...
Equation of tangent and normal at point P(at2, 2at) is
ty = x + at2 and y = −tx + 2at + at2
Let the centroid of △PTN be R(h, k).
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The tangent PT and the normal PN to the parabola y2 = 4ax at a point P on it meet its axis at points T and N, respectively. The locus of the centroid of the triangle PTN is a parabola whosea)vertex is (2a/3 , 0)b)directrix is x = 0c)latus rectum is 2a/3d)focus is (a, 0)Correct answer is option 'A,D'. Can you explain this answer?
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The tangent PT and the normal PN to the parabola y2 = 4ax at a point P on it meet its axis at points T and N, respectively. The locus of the centroid of the triangle PTN is a parabola whosea)vertex is (2a/3 , 0)b)directrix is x = 0c)latus rectum is 2a/3d)focus is (a, 0)Correct answer is option 'A,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 The tangent PT and the normal PN to the parabola y2 = 4ax at a point P on it meet its axis at points T and N, respectively. The locus of the centroid of the triangle PTN is a parabola whosea)vertex is (2a/3 , 0)b)directrix is x = 0c)latus rectum is 2a/3d)focus is (a, 0)Correct answer is option 'A,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 The tangent PT and the normal PN to the parabola y2 = 4ax at a point P on it meet its axis at points T and N, respectively. The locus of the centroid of the triangle PTN is a parabola whosea)vertex is (2a/3 , 0)b)directrix is x = 0c)latus rectum is 2a/3d)focus is (a, 0)Correct answer is option 'A,D'. Can you explain this answer?.
The tangent PT and the normal PN to the parabola y2 = 4ax at a point P on it meet its axis at points T and N, respectively. The locus of the centroid of the triangle PTN is a parabola whosea)vertex is (2a/3 , 0)b)directrix is x = 0c)latus rectum is 2a/3d)focus is (a, 0)Correct answer is option 'A,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 The tangent PT and the normal PN to the parabola y2 = 4ax at a point P on it meet its axis at points T and N, respectively. The locus of the centroid of the triangle PTN is a parabola whosea)vertex is (2a/3 , 0)b)directrix is x = 0c)latus rectum is 2a/3d)focus is (a, 0)Correct answer is option 'A,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 The tangent PT and the normal PN to the parabola y2 = 4ax at a point P on it meet its axis at points T and N, respectively. The locus of the centroid of the triangle PTN is a parabola whosea)vertex is (2a/3 , 0)b)directrix is x = 0c)latus rectum is 2a/3d)focus is (a, 0)Correct answer is option 'A,D'. Can you explain this answer?.
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