Write the equation for a parabola with a focus at (-6,0) and a directrix at x=-2
***PLEASE SET AS "x="***
sqrt[(x+2)^2] = sqrt[(x+6)^2 + y^2]
x^2+4x+4=x^2+12x+36+y^2
4x-12x=y^2=36-4
-8x=y^2+32
x=(-y^2/8)-4
Sure thing! I'll set it as "x=" for you.
The equation for a parabola with a focus at (-6,0) and a directrix at x=-2 can be written as:
(x + 6)^2 = 4p(y - 0)
Since the directrix is x = -2, the distance between the focus and directrix is the same as the distance between the vertex and directrix. We denote this distance as 2p. So, in this case, 2p = 4.
Substituting this value into the equation, we have:
(x + 6)^2 = 16(y - 0)
Simplifying further, we get:
x + 6 = ±4√(y - 0)
And setting it as "x=", we have:
x = ±4√(y - 0) - 6
So the equation of the parabola is x = ±4√y - 6.
To find the equation for a parabola with a focus and directrix, we need to use the standard form of the equation for a parabola:
(x - h)² = 4p(y - k)
Where (h, k) represents the vertex of the parabola, p represents the distance between the vertex and the focus (or the vertex and the directrix), and (x, y) represents any point on the parabola.
In this case, the focus is at (-6, 0), which means the vertex is also at (-6, 0) since focus and vertex are at the same x-coordinate. Additionally, the directrix is at x = -2, so the distance between the vertex and the directrix (p) is 4 units (since it's the absolute difference between the x-coordinate of the vertex and the x-coordinate of the directrix).
Using these values, we can substitute them into the standard form equation:
(x - (-6))² = 4(4)(y - 0)
Simplifying:
(x + 6)² = 16y
Therefore, the equation of the parabola with focus at (-6, 0) and directrix at x = -2 is:
x + 6 = ±√(16y)
Hgff
Revie the parabola y^2 = 4px
Yours is
y^2 = -8(x+4)