770 Words4 Pages

The parabola
Let us start the lesson by knowing the definition of the different terms or key words which are necessary to understand more the concepts in this lesson.
Activity 1.2.a. Tell me More... Define each of the following terms or draw a figure illustrating these terms:
Vertex
Curve opens upward
Curve opens downward
Concavity of a curve
Axis of symmetry
The Parabola and its parts
A parabola is a locus of points P(x,y) in a plane equidistant from a fixed point and a fixed line. The fixed point of a parabola is the focus and the fixed line of a parabola is known as the directrix. The vertex is the turning point of the curve. The distance from the focus to the vertex or the distance from*…show more content…*

The line y = − p is the directrix The vertex is at the origin (0, 0). The focal distance is ∣p∣ The latus rectum is AB and its length is 4p. The axis of symmetry is on the y-axis or a line passing through the vertex perpendicular to the latus rectum. The point P(x, y) is any point on the curve. The distance from the point (x, y) to the directrix is the same from the distance from any point (x, y) to the focus (0, p) The parabola opens upward. Activity 1.2.b. Step-by-Step Consider the parabola below. Derive the equation of a parabola whose vertex is at the origin and opens to the right by following the given steps. Based from the definition of the parabola: x + p = ⃒ PF⃒ Substitute the value of ⃒ PF⃒ by its length d using the distance formula: d=x+p=√((〖x-x〗_1 )^2+(〖y-y〗_1 )^2 ) Substitute the values x1 = p and y1 = 0: x+p=√((x-p)^2+(y-0)^2 ) Square both sides of the equation: (x+p)^2=(x-p)^2+y^2 Expand the squares of binomials: x^2+2xp+p^2=x^2-2xp+p^2+y^2 Simplify by adding or subtracting common terms:*…show more content…*

Part-by-Part Study each part of the parabola above by answering the following questions below: Determine the equation of the parabola The parabola opens to what direction? Find the distance of the vertex to the focal point of the parabola. Solve the length of the latus rectum. Give the equation of the directrix. Case I. Vertex at Origin Parabola with Vertical Axis Adding to our diagram, we see that the distance d = y + p. Figure 1.2.2 Now, using the Distance Formula on the general points (0, p) and (x, y), and equating it to our value d = y + p, we have d = √(〖(x_2- x_1)〗^2+ (〖y_2- y_1)〗^2 ) y + p = √(〖(x- 0)〗^2+ (〖y- p)〗^2 ) Squaring both sides gives: (y + p)2 = (x − 0)2 + (y − p)2 y2 + 2py + p2 = x2 + y2 – 2py + p2 Simplifying gives us the formula for a parabola: x2 = 4py In more familiar form, with "y = " on the left, we can write this as: y = x^2/4p where p is the focal distance of the parabola. Example 1.1.a. Sketch the parabola x2 = 4y. Find the focal length and indicate the focus and the directrix on your

The line y = − p is the directrix The vertex is at the origin (0, 0). The focal distance is ∣p∣ The latus rectum is AB and its length is 4p. The axis of symmetry is on the y-axis or a line passing through the vertex perpendicular to the latus rectum. The point P(x, y) is any point on the curve. The distance from the point (x, y) to the directrix is the same from the distance from any point (x, y) to the focus (0, p) The parabola opens upward. Activity 1.2.b. Step-by-Step Consider the parabola below. Derive the equation of a parabola whose vertex is at the origin and opens to the right by following the given steps. Based from the definition of the parabola: x + p = ⃒ PF⃒ Substitute the value of ⃒ PF⃒ by its length d using the distance formula: d=x+p=√((〖x-x〗_1 )^2+(〖y-y〗_1 )^2 ) Substitute the values x1 = p and y1 = 0: x+p=√((x-p)^2+(y-0)^2 ) Square both sides of the equation: (x+p)^2=(x-p)^2+y^2 Expand the squares of binomials: x^2+2xp+p^2=x^2-2xp+p^2+y^2 Simplify by adding or subtracting common terms:

Part-by-Part Study each part of the parabola above by answering the following questions below: Determine the equation of the parabola The parabola opens to what direction? Find the distance of the vertex to the focal point of the parabola. Solve the length of the latus rectum. Give the equation of the directrix. Case I. Vertex at Origin Parabola with Vertical Axis Adding to our diagram, we see that the distance d = y + p. Figure 1.2.2 Now, using the Distance Formula on the general points (0, p) and (x, y), and equating it to our value d = y + p, we have d = √(〖(x_2- x_1)〗^2+ (〖y_2- y_1)〗^2 ) y + p = √(〖(x- 0)〗^2+ (〖y- p)〗^2 ) Squaring both sides gives: (y + p)2 = (x − 0)2 + (y − p)2 y2 + 2py + p2 = x2 + y2 – 2py + p2 Simplifying gives us the formula for a parabola: x2 = 4py In more familiar form, with "y = " on the left, we can write this as: y = x^2/4p where p is the focal distance of the parabola. Example 1.1.a. Sketch the parabola x2 = 4y. Find the focal length and indicate the focus and the directrix on your

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