parabola with vertex at (3,-2) and a directreix of x=2. how do i wrtie this as an equation....PLEASE HELPPPPPPPP
Let's use basic concepts for this one
let P(x,y) be any point on the parabola
the focal point will be F(4,-2), (remember the vertex is midway between the focal point and the directrix)
Then PF = distance from P to directrix
√((x-4)^2+(y+2)^2) = √(x-2)^2
square both sides and expand
x^2 - 8x + 15 + y^2 + 4y + 4 = x^2 - 4x + 4
y^2 + 4y + 20 = 4x
from here you could complete the square to get
x = 1/4(y+2)^2 + 3 to confirm that the vertex is (3,-2)
To write the equation of a parabola with a vertex at (h, k) and a directrix of x = d, you can use the following standard form equation:
(x - h)^2 = 4p(y - k)
In this case, the vertex is (3, -2) and the directrix is x = 2.
Step 1: Find the value of p.
The distance between the vertex and the directrix is equal to the distance between the vertex and the focus, and it's denoted as p. In this case, the vertex is (3, -2), and the directrix is x = 2. The distance between them is 3 - 2 = 1, so p = 1.
Step 2: Substitute the values into the standard form equation.
Substituting h = 3, k = -2, and p = 1 into the equation:
(x - 3)^2 = 4(1)(y - (-2))
(x - 3)^2 = 4(y + 2)
Therefore, the equation of the parabola with a vertex at (3, -2) and a directrix of x = 2 is (x - 3)^2 = 4(y + 2).
Please note that this equation represents a vertical parabola. If the parabola is horizontal, the standard form equation would be (y - k)^2 = 4p(x - h), and you would substitute the given values accordingly.