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The slope of the normal at the point (at2,2at) of the parabola y2=4ax is
- a)1/t
- b)t
- c)-t
- d)-1/t
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
The slope of the normal at the point (at2,2at) of the parabola y2=4ax ...
Understanding the Parabola
The equation of the parabola is given as y² = 4ax. Here, the focus is at (a, 0) and the directrix is x = -a.
Point on the Parabola
We are examining the point (at², 2at). To verify that this point lies on the parabola, substitute x = at² into the parabola's equation:
- Substitute: (2at)² = 4a(at²)
- Simplifies to: 4a²t² = 4a²t², confirming the point is on the parabola.
Finding the Slope of the Tangent
To find the slope of the tangent at the point (at², 2at):
- Differentiate y² = 4ax implicitly:
- 2y(dy/dx) = 4a
- Hence, dy/dx = 4a/(2y) = 2a/y
- Substitute y = 2at:
- dy/dx = 2a/(2at) = 1/t
Slope of the Normal
The normal line is perpendicular to the tangent. Hence, the slope of the normal is the negative reciprocal of the slope of the tangent:
- Slope of the normal = -1/(slope of the tangent) = -1/(1/t) = -t
Conclusion
Therefore, the slope of the normal at the point (at², 2at) on the parabola y² = 4ax is:
- Correct Answer: c) -t
The equation of the parabola is given as y² = 4ax. Here, the focus is at (a, 0) and the directrix is x = -a.
Point on the Parabola
We are examining the point (at², 2at). To verify that this point lies on the parabola, substitute x = at² into the parabola's equation:
- Substitute: (2at)² = 4a(at²)
- Simplifies to: 4a²t² = 4a²t², confirming the point is on the parabola.
Finding the Slope of the Tangent
To find the slope of the tangent at the point (at², 2at):
- Differentiate y² = 4ax implicitly:
- 2y(dy/dx) = 4a
- Hence, dy/dx = 4a/(2y) = 2a/y
- Substitute y = 2at:
- dy/dx = 2a/(2at) = 1/t
Slope of the Normal
The normal line is perpendicular to the tangent. Hence, the slope of the normal is the negative reciprocal of the slope of the tangent:
- Slope of the normal = -1/(slope of the tangent) = -1/(1/t) = -t
Conclusion
Therefore, the slope of the normal at the point (at², 2at) on the parabola y² = 4ax is:
- Correct Answer: c) -t
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The slope of the normal at the point (at2,2at) of the parabola y2=4ax ...
Understanding the Parabola
The equation of the parabola given is y² = 4ax. In this case, a is a constant that determines the width of the parabola.
Point on the Parabola
We have a specific point on the parabola represented by (at², 2at). Here, t is a parameter that can vary.
Finding the Slope of the Tangent
1. To find the slope of the tangent at the given point (at², 2at):
- Differentiate y² = 4ax implicitly with respect to x.
- This gives us dy/dx = 2a/y.
- At the point (at², 2at), substituting y = 2at gives:
- dy/dx = 2a/(2at) = a/t.
Slope of the Normal Line
2. The slope of the normal line is the negative reciprocal of the slope of the tangent:
- Slope of normal = -1/(dy/dx).
- Therefore, substituting dy/dx = a/t gives:
- Slope of normal = -t/a.
However, we also need to consider that the slope of the tangent line is actually a/t, so the negative reciprocal becomes -t.
Conclusion
Thus, the slope of the normal at the point (at², 2at) is indeed -t. The correct answer is option 'C' (-t), which is consistent with our calculations and understanding of the relationship between slopes of tangents and normals.
The equation of the parabola given is y² = 4ax. In this case, a is a constant that determines the width of the parabola.
Point on the Parabola
We have a specific point on the parabola represented by (at², 2at). Here, t is a parameter that can vary.
Finding the Slope of the Tangent
1. To find the slope of the tangent at the given point (at², 2at):
- Differentiate y² = 4ax implicitly with respect to x.
- This gives us dy/dx = 2a/y.
- At the point (at², 2at), substituting y = 2at gives:
- dy/dx = 2a/(2at) = a/t.
Slope of the Normal Line
2. The slope of the normal line is the negative reciprocal of the slope of the tangent:
- Slope of normal = -1/(dy/dx).
- Therefore, substituting dy/dx = a/t gives:
- Slope of normal = -t/a.
However, we also need to consider that the slope of the tangent line is actually a/t, so the negative reciprocal becomes -t.
Conclusion
Thus, the slope of the normal at the point (at², 2at) is indeed -t. The correct answer is option 'C' (-t), which is consistent with our calculations and understanding of the relationship between slopes of tangents and normals.
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