Inactive Tutor answered • 04/21/23
x=-6 is the directrix
focus is (-10,5)
it's a leftward opening horizontal parabola
a quadratic equation of the form x=a(y-k)^2 + h
where (h,k) is the vertex, the most rightward point on the parabola
vertex is midway the midpoint between the focus and direction, with y coordinate same as the focus
vertex = (-8,5) -8 is the average of -10 and -6
x = a(y-5)^2 -8
all you need now is to find the value of "a," the coefficient of the x^2 term, the compression or stretching factor
find any other point on the parabola, plug it in and solve for "a"
one point is directly above or below the focus with x coordinate = -10
and y coordinate is found by setting the distance from focus to parabola = distance from parabola to directrix
from y = 5 to y coordinate of the parabola = from x coordinate of the parabola to x=-10
3 points on the parabola are (-8,5), (-10,9) and (-10,1). last 2 points are the endpoints of the latus rectum
plug in either of the last two points to find "a"
x=a(y-k)^2 +h
x =a(y-5)^2 -8
-10= 16a -8
16a =-2
a=-2/16=-1/8
x = (-1/8)(y-5)^2 -8
or a = -1/4p
p=h--10=-8+10=2
-1/4p=-1/4(2) = -8
x = (-1/8)(y-5)^2 -8
if you work a problem 2 different ways and get the same answer, it's very likely you're right
"a" should be negative,
x = (-1/8)(y-5)^2 -8 is the equation of a parabola with directrix x=-6 and focus (-10, 5)
it helps to graph the directrix line, and the points, focus, vertex and points directly above the focus on the parabola, get a better feel for how things look
