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Any point on the parabola whose focus is (0,1) and the directrix is x + 2 = 0 is given by
a)(t2 + 1, 2t – 1)
b)(t2, 2t)
c)(t2 + 1, 2t + 1)
d)(t2 – 1, 2t + 1)?
a)(t2 + 1, 2t – 1)
b)(t2, 2t)
c)(t2 + 1, 2t + 1)
d)(t2 – 1, 2t + 1)?
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Any point on the parabola whose focus is (0,1) and the directrix is x ...
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Any point on the parabola whose focus is (0,1) and the directrix is x ...
Introduction:
In this problem, we are given a parabola with a focus at (0,1) and a directrix given by x-2=0. We need to find the coordinates of any point on the parabola.
Understanding the Parabola:
To understand the parabola, let's first define its key elements:
- Focus: The focus of a parabola is a fixed point on the interior of the parabola. In this case, the focus is given as (0,1).
- Directrix: The directrix of a parabola is a fixed line outside the parabola. In this case, the directrix is given as x-2=0.
Definition of a Parabola:
A parabola is the set of all points that are equidistant from the focus and the directrix.
Finding the Equation of the Parabola:
To find the equation of the parabola, we need to determine the vertex and the focal length.
Finding the Vertex:
The vertex of a parabola is the point halfway between the focus and the directrix. Since the directrix is a vertical line, the x-coordinate of the vertex is the same as the x-coordinate of the focus. Therefore, the vertex is (0,1).
Finding the Focal Length:
The focal length is the distance between the focus and the vertex. In this case, the focal length is 1 unit.
Finding the Equation of the Parabola:
The equation of a parabola with a vertical axis of symmetry can be written as (x-h)^2 = 4p(y-k), where (h,k) is the vertex and p is the focal length.
Substituting the values, we have (x-0)^2 = 4(1)(y-1), which simplifies to x^2 = 4(y-1).
Interpreting the Equation:
From the equation, we can see that the parabola opens upwards and its vertex is at (0,1). The coefficient of x^2 is positive, indicating that the parabola is facing upwards.
Finding the Coordinates of any Point on the Parabola:
To find the coordinates of any point on the parabola, we can use the parameterization of the parabola in terms of a parameter t.
Let's assume the x-coordinate of the point is t. Then the corresponding y-coordinate can be found by substituting t into the equation of the parabola.
Substituting t into the equation x^2 = 4(y-1), we have t^2 = 4(y-1).
Simplifying the equation, we get y = (t^2 + 1)/4.
Therefore, the coordinates of any point on the parabola are (t, (t^2 + 1)/4).
Answer:
The correct answer is (t^2, 2t) given by option (b) because it matches the derived equation for the coordinates of any point on the parabola. Option (b) represents the x-coordinate as t^2 and the y-coordinate as 2t, which is equivalent to (t, (t^2 +
In this problem, we are given a parabola with a focus at (0,1) and a directrix given by x-2=0. We need to find the coordinates of any point on the parabola.
Understanding the Parabola:
To understand the parabola, let's first define its key elements:
- Focus: The focus of a parabola is a fixed point on the interior of the parabola. In this case, the focus is given as (0,1).
- Directrix: The directrix of a parabola is a fixed line outside the parabola. In this case, the directrix is given as x-2=0.
Definition of a Parabola:
A parabola is the set of all points that are equidistant from the focus and the directrix.
Finding the Equation of the Parabola:
To find the equation of the parabola, we need to determine the vertex and the focal length.
Finding the Vertex:
The vertex of a parabola is the point halfway between the focus and the directrix. Since the directrix is a vertical line, the x-coordinate of the vertex is the same as the x-coordinate of the focus. Therefore, the vertex is (0,1).
Finding the Focal Length:
The focal length is the distance between the focus and the vertex. In this case, the focal length is 1 unit.
Finding the Equation of the Parabola:
The equation of a parabola with a vertical axis of symmetry can be written as (x-h)^2 = 4p(y-k), where (h,k) is the vertex and p is the focal length.
Substituting the values, we have (x-0)^2 = 4(1)(y-1), which simplifies to x^2 = 4(y-1).
Interpreting the Equation:
From the equation, we can see that the parabola opens upwards and its vertex is at (0,1). The coefficient of x^2 is positive, indicating that the parabola is facing upwards.
Finding the Coordinates of any Point on the Parabola:
To find the coordinates of any point on the parabola, we can use the parameterization of the parabola in terms of a parameter t.
Let's assume the x-coordinate of the point is t. Then the corresponding y-coordinate can be found by substituting t into the equation of the parabola.
Substituting t into the equation x^2 = 4(y-1), we have t^2 = 4(y-1).
Simplifying the equation, we get y = (t^2 + 1)/4.
Therefore, the coordinates of any point on the parabola are (t, (t^2 + 1)/4).
Answer:
The correct answer is (t^2, 2t) given by option (b) because it matches the derived equation for the coordinates of any point on the parabola. Option (b) represents the x-coordinate as t^2 and the y-coordinate as 2t, which is equivalent to (t, (t^2 +
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Any point on the parabola whose focus is (0,1) and the directrix is x + 2 = 0 is given bya)(t2 + 1, 2t – 1)b)(t2, 2t)c)(t2 + 1, 2t + 1)d)(t2 – 1, 2t + 1)?
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Any point on the parabola whose focus is (0,1) and the directrix is x + 2 = 0 is given bya)(t2 + 1, 2t – 1)b)(t2, 2t)c)(t2 + 1, 2t + 1)d)(t2 – 1, 2t + 1)? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Any point on the parabola whose focus is (0,1) and the directrix is x + 2 = 0 is given bya)(t2 + 1, 2t – 1)b)(t2, 2t)c)(t2 + 1, 2t + 1)d)(t2 – 1, 2t + 1)? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Any point on the parabola whose focus is (0,1) and the directrix is x + 2 = 0 is given bya)(t2 + 1, 2t – 1)b)(t2, 2t)c)(t2 + 1, 2t + 1)d)(t2 – 1, 2t + 1)?.
Any point on the parabola whose focus is (0,1) and the directrix is x + 2 = 0 is given bya)(t2 + 1, 2t – 1)b)(t2, 2t)c)(t2 + 1, 2t + 1)d)(t2 – 1, 2t + 1)? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Any point on the parabola whose focus is (0,1) and the directrix is x + 2 = 0 is given bya)(t2 + 1, 2t – 1)b)(t2, 2t)c)(t2 + 1, 2t + 1)d)(t2 – 1, 2t + 1)? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Any point on the parabola whose focus is (0,1) and the directrix is x + 2 = 0 is given bya)(t2 + 1, 2t – 1)b)(t2, 2t)c)(t2 + 1, 2t + 1)d)(t2 – 1, 2t + 1)?.
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