Thanks to those who helped with my question: y+3=-1/12(x-1)^2. The only thing I am still unsure of is how to write the equation for the directrix. Do I just simply write 0=3-3, (the values plugged in for y=k-p). Thanks again for your help!!
3 answers
Whoops and one thing I forgot. Is it allowed for a parobala to open upward and have the directrix intercept it? Since p is greater than zero, it would have to open upward, but then the directrix would intercept the parobala (the directrix follows the x axis, and is at zero y)
p = Negative. (I called p a but same difference)
I showed you that this parabola opens downward in my explanation of the axis of symmetry at x = 1
the directrix is a horizontal line (m=0)
y = 0 x + 0
or
y = 0
I showed you that this parabola opens downward in my explanation of the axis of symmetry at x = 1
the directrix is a horizontal line (m=0)
y = 0 x + 0
or
y = 0
The form is
(x-h)^2 = 4 a (y-k)
here
(x-1)^2 = -12 (y+3)
so 4 a = -12
a = -3
vertex to focus = a = -3
so focus at -3-3 = -6 so (1,-6)
directrix is horizontal line a units above vertex so it is the line y = 0
(x-h)^2 = 4 a (y-k)
here
(x-1)^2 = -12 (y+3)
so 4 a = -12
a = -3
vertex to focus = a = -3
so focus at -3-3 = -6 so (1,-6)
directrix is horizontal line a units above vertex so it is the line y = 0
1 answer