# Isaiah's Math Lessons : Parabola

• 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)?

1.     x^2 + 2x - 3y - 2 = 0

2.     x^2 - x + y = 0

1

1

1

1

1

1

1

1