SOLUTION: How do you find the vertex, value of p, axis of symmetry, focus, and directrix of each parabola with the equation X-1=-1/12y^2

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: How do you find the vertex, value of p, axis of symmetry, focus, and directrix of each parabola with the equation X-1=-1/12y^2      Log On


   

Question 997627: How do you find the vertex, value of p, axis of symmetry, focus, and directrix of each parabola with the equation X-1=-1/12y^2
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
The equation written correctly, or in more precise manner is x-1=-%281%2F12%29y%5E2.
This is a parabola with symmetry axis parallel to the x-axis, and graph of parabola opens to the left, as indicated by coefficient NEGATIVE 1/12. This coefficient -1%2F12 tells you information about how far is focus and directrix from the vertex. YOUR equation's vertex is (1,0).

Learn better how this works through these two videos:

Derive equation for parabola from focus and directrix

Equation for parabola from directrix and focus but vertex is NOT at the Origin

According to the form of the equation used in the derivations, 4p=%281%2F12%29, and |p| is how far vertex is from either focus or directrix.
-
p=%281%2F4%29%281%2F12%29
p=1%2F48

Focus is on the concave side, to left of the vertex.
x=1-1%2F48
x=48%2F48-1%2F48
x=47%2F48 and y=0, this y-value unchanged.
FOCUS: ( 47/48, 0 )

Directrix would be on the other side and is the vertical line x=1%261%2F48.