Isaiah's Math Lessons : Parabola
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Parabola Definition:
The locus of a point that moves it is equidistant from a fixed point called focus and a fixed line called directrix.
General Equation:
Ax^2 + Dx + Ey + F = 0 (Axis Parallel to the y –axis)
Ay^2 + Dx + Ey + F = 0 (Axis Parallel to the x –axis)
Standard Equation:
Parabola with vertex @ (h, k)
(y-k)^2 = 4a (x-h) --- > Opens to the right
(y-k)^2 = -4a (x-h) -- > Opens to the left
(x-h)^2 = 4a (y-k) --- > Opens upward
(x-h)^2 = - 4a (y-k) --- > Opens downward
KEY FORMULAS:
4a = Length of Latus Rectum (LR)
a = focal distance
(h + a, k) = coordinates of the focus if axis is parallel to the x - axis
(h, k + a) = coordinates of the focus if axis is parallel to the y - axis
(h + a, k +/- 2a) = coordinates of the endpoints of LR if axis is parallel to the x – axis
(h +/- 2a, k + a) = coordinates of the endpoints of LR if axis is parallel to the y - axis
e = 1 (always constant for parabola)
Examples and Detailed Solutions:
1. Find the general equation of the parabola with focus at (1, 0) and vertex at (2, 0).
Solution:
By plotting the points, we can directly say that the parabola is opening to the left or with axis parallel to the x-axis.
Thus a = 1.
(y-k)^2 = -4a(x-h)
y^2 = -4(1)(x-2)
y^2 = -4(x-2)
y^2 + 4x - 8 = 0
2. Find the general equation of the parabola with vertex at (0, 3) and directrix at x = -1.
Solution:
By graphing and plotting the points, this parabola has an axis of symmetry parallel to the y-axis.
Distance from directrix and vertex is equal to a. Thus, a = 1 and LR = 4. Also, by inspection, the graph opens to the right.
(y-k)^2 = 4a(x-h)
(y-3)^2 = 4(1)(x)
y^2 - 4x - 6y + 9 = 0
Practice Problems:
1. Find the equation of the parabola with axis vertical, vertex at (-1, -1) and passing through (2, 2).
2. What is the equation of the parabola with axis vertical and passing through (0, 0), (1, 0) and (5, -20)?
Answers:
1. x^2 + 2x - 3y - 2 = 0
2. x^2 - x + y = 0
