Posts Tagged ‘vertex’

Problem 1084

Find the vertex, focus, and directrix of the following parabola. Then draw the graph.

(x – 2)2 = -3 (y + 1)




The two standard forms of the equation for a parabola are (x – h)2 = 4p(y – k) or

(y – k)2 = 4p(x – k).

Since the parabola is already in one of the standard forms, you can determine the values of h, k, and p, and form these values determine the vertex, focus, and directrix.

h = 2, k = -1, and p = –\frac{3}{4}

Remember, for a parabola written in the form (x – h)2 = 4p(y – k), the vertex is (h , k).

Thus , the vertex of this parabola is (2,-1).

The axis of symmetry is vertical. The focus is –\frac{3}{4} units away from the vertex, in the direction of the axis of symmetry.

Thus, it will be \frac{3}{4} units below the vertex.

The focus is (2,-\frac{7}{4}).

The directrix will be \frac{3}{4} units above the vertex.

The equation for the directrix is

y = –\frac{1}{4}

With this information, graph the parabola. First, graph the vertex, focus, and axis of symmetry.

Next, fill in the graph of the parabola.



Problem 1081

Find the vertex, the focus, and the directrix.  Then draw the graph.

y2 = -6x




An equation of the form y2 = 4px is the standard equation of a parabola with the vertex at the origin, focus at (p,0) and directrix x = -p.

To write the equation y2 = -6x in the form y2 = 4px, p = –\frac{3}{2}

Thus , y2 = -6x is equivalent to y2 = 4(-\frac{3}{2})x.

The vertex is located at (0,0).

The focus is at (-\frac{3}{2},0).

The directrix is the line x=\frac{3}{2}.

The correct graph of the relation is shown to the right.