How do you graph a parabola with only the vertex and directrix given?
I've already plotted the coordinates of the vertex and the directrix. I just don't completely understand how I could get the focus.
1) Vertex : (8, 3)
Directrix : y = 7
2) Vertex : (0, 3)
Directrix : x = 4
- RaffaeleLv 76 years agoFavorite Answer
look here for a diagram
The focus F is on the symmetry axes of the parabola x = 8
V(8,3) is 4 far from the directrix (7 - 3 = 4)
FV = 4
remember that F is "inside" the parabola
you can find the equation directly
P(x,y) is a point of the parabola
PF = PH (where H is the projection of P on the directrix)
PF² = PH²
(x - 8)² + (y + 1)² = (y - 7)²
y = - x^2/16 + x - 1
x = 4
symmetry axes y = 3
(x + 4)² + (y - 3)² = (x - 4)²
x = - y^2/16 + (3/8)y - 9/16
- ignoramusLv 76 years ago
I presume that you know that the equation of a parabola whose vertex is at the origin, and whose axis is the positive x-axis is
y² = 4ax
Since the definition of a parabola is that any point on the parabola is the same distance from the focus and the directrix, it follows that the focus is the same distance from the vertex (but on the opposite side) as is the directrix. The parameter a in the equation is the distance from the vertex to the directrix (on one side), and to the focus on the other side. That is, if the equation is y² = 4x, then the directrix is at y = -1, and the focus is at (1, 0).
1) Since the directrix is y = 7, and the vertex is at x = 8, it means that the focus must be at x = 9. The directrix is vertical, therefore the axis is horizontal, therefore the focus is at (9, 3)
2) Similaly, for this one, the directrix being to the right of the vertex means that the parobola opens to the left, so the focus must be at x = -4, that is, the point (-4, 3).