Isaiah's Math Lessons : Parabola
isaiah_james last edited by isaiah_james
The locus of a point that moves it is equidistant from a fixed point called focus and a fixed line called directrix.
Ax^2 + Dx + Ey + F = 0 (Axis Parallel to the y –axis)
Ay^2 + Dx + Ey + F = 0 (Axis Parallel to the x –axis)
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
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).
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.
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
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)?
1. x^2 + 2x - 3y - 2 = 0
2. x^2 - x + y = 0