Question

Which is the directrix of a parabola with equation mc001-1.jpg?
mc001-2.jpg
mc001-3.jpg
mc001-4.jpg
mc001-5.jpg

Answer 1

I can solve it, but I can't see the images you attached. Please either paste the parabola equation and the answer choices as text, or re-upload the images (or describe them).

Meanwhile, here is how to find the directrix so you can check quickly:

- If the parabola is in standard vertical form (x − h)^2 = 4p(y − k), its vertex is (h,k), focus is (h, k + p) and the directrix is y = k − p.
- If it is in standard horizontal form (y − k)^2 = 4p(x − h), its vertex is (h,k), focus is (h + p, k) and the directrix is x = h − p.
- If you have y = a(x − h)^2 + k (that is, vertical parabola in vertex form), then 4p = 1/a so p = 1/(4a) and the directrix is y = k − p.
- If you have a general quadratic y = ax^2 + bx + c, complete the square to get y = a(x − h)^2 + k, then use the formula above.
- If the equation has an xy term (rotation), the parabola is rotated and you must rotate axes (or use conic classification) to find the directrix.

Example: y = 2x^2 + 4x + 1
- Complete the square: y = 2[(x+1)^2 − 1] + 1 = 2(x+1)^2 − 1
- Vertex: (−1, −1); a = 2 → p = 1/(4a) = 1/8
- Directrix: y = k − p = −1 − 1/8 = −9/8

Send the equation or choices and I’ll pick the correct directrix.