Question

Which is the directrix of a parabola with equation mc002-1.jpg?
mc002-2.jpg
mc002-3.jpg
mc002-4.jpg
mc002-5.jpg

Answer 1

I can’t view the image files you attached. Please either paste the parabola equation here or retype it (or upload a single image of the equation).

Meanwhile, here are quick rules you can use to find the directrix:

1. Put the equation into one of the standard forms by completing the square:
- Vertical parabola: (x − h)^2 = 4p(y − k). Focus = (h, k + p). Directrix: y = k − p.
- Horizontal parabola: (y − k)^2 = 4p(x − h). Focus = (h + p, k). Directrix: x = h − p.

2. If your equation is y = a(x − h)^2 + k, then compare y − k = a(x − h)^2 with y − k = (1/(4p))(x − h)^2, so p = 1/(4a). Directrix: y = k − p.

3. If the general second-degree form has an xy-term (B ≠ 0), a rotation is needed before converting to standard form.

Paste the equation and I’ll compute the directrix for you.