SOLUTION: Find the equation in standard form of the parabola with directrix x = 3 and vertex (1, -2)

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Find the equation in standard form of the parabola with directrix x = 3 and vertex (1, -2)      Log On


   



Question 766994: Find the equation in standard form of the parabola with directrix x = 3 and vertex (1, -2)
Found 2 solutions by lwsshak3, josgarithmetic:
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
Find the equation in standard form of the parabola with directrix x = 3 and vertex (1, -2)
Equation is that of a parabola which open leftward:
Its basic equation: (y-k)^2=-4p(x-h), (h,k)=(x,y) coordinates of the vertex.
For given parabola:
vertex:(1,-2)
axis of symmetry: y=-2
p=3 (distance from vertex to directrix on the axis of symmetry)
4p=12
Equation of given parabola: (y+2)^2=-12(x-1)

Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
Try first putting points onto graph paper. Focus on other side of vertex equally from the directrix. Focus (-1,-2). The picture also helps you label the curve and general point (x,y), and set up distance equation knowing the general point is the same distance from focus (-1,-2) as it is from directrix (3,y).

I suggest doing the steps starting from

simply becasue putting all the text steps here will be difficult, in contrast to doing it on paper.
.
Completing The Square was also needed.
.
After the steps and simplifying, you should get something equivalent to

highlight%28-%281%2F8%29%28y%2B2%29%5E2=x%2B1%29