Identify th equation of the parabola whose focus is at (5,-1) and whose directrix is x=3
I don't see any examples of this kind of problem.
is this right?
(x-5)^2+(y+1)^2=(x-3)^2
so
-10x+25+y^2+2y+1=-6x+9
4x=y^2+2y+17
http://home.alltel.net/okrebs/page64.html
see problem six.
vertex=(-5,-1)
p=-6
4p=-24
(x-5)=(-24)(y+1)^2
correction (y+3)^2
To identify the equation of a parabola given its focus and directrix, you can use the definition of a parabola as a set of points equidistant from the focus and the directrix. Let's break down the steps to find the equation:
1. Start by noting the coordinates of the focus and the equation of the directrix.
- Focus: (5, -1)
- Directrix: x = 3
2. Use the distance formula to express the distance between a point (x, y) on the parabola to the focus (5, -1) and the vertical directrix x = 3.
- Distance from a point to the focus: √((x - 5)^2 + (y - (-1))^2)
- Distance from a point to the directrix: |x - 3|
3. Set the two distances equal to each other to find the equation of the parabola.
√((x - 5)^2 + (y + 1)^2) = |x - 3|
4. Square both sides of the equation to eliminate the square root and simplify.
(x - 5)^2 + (y + 1)^2 = (x - 3)^2
It seems like you made a slight error in your calculation. Let's correct it:
Expanding (x - 5)^2, (y + 1)^2, and (x - 3)^2, we get:
x^2 - 10x + 25 + y^2 + 2y + 1 = x^2 - 6x + 9
Simplifying this equation, we have:
-10x + y^2 + 2y + 26 = -6x + 9
Rearranging the terms to get the equation in standard form, we get:
-10x + 6x + y^2 + 2y = 9 - 26
-4x + y^2 + 2y = -17
So, the corrected equation of the parabola with a focus at (5, -1) and a directrix x = 3 is:
-4x + y^2 + 2y = -17