Ask Question
20 December, 13:03

Find the x‑ and y‑coordinates of the focus of the parabola that is represented by the equation below. Enter your answers as integers. y=0.125x2+4x-29 Focus

+2
Answers (1)
  1. 20 December, 13:22
    0
    In form

    4p (y-k) = (x-h) ^2

    vertex = (h, k)

    p is the horizontal distance from the vertex to the focus and directix

    when p is positive, the focus is above the vertex

    complete the square

    y = (0.125x^2+4x) - 29

    y=0.125 (x^2+32x) - 29

    32/2=16, 16^2=256

    add negative and positive inside

    y=0.125 (x^2+32x+256-256) - 29

    complete the square

    y=0.125 ((x+16) ^2-256) - 29

    distribute

    y=0.125 (x+16) ^2-32-29

    y=0.125 (x+16) ^2-61

    add 61 to both sides

    y+61=0.125 (x+16) ^2

    divide both sides by 0.125

    8 (y+61) = (x+16) ^2

    4 (2) (y+61) = (x+16) ^2

    4 (2) (y - (-61)) = (x - (-16)) ^2

    4p (y-k) = (x-h) ^2

    vertex at (-16,-61)

    p is 2 which is positive so it s above

    -61+2=-59

    focus at (-16,-59)
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “Find the x‑ and y‑coordinates of the focus of the parabola that is represented by the equation below. Enter your answers as integers. ...” in 📙 Mathematics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers