Question

Finding the vertex focus directrix and axis of symmetry of a parabola

Answer 1

Equation:

$(y+1)^2=6(x-5)$

The vertex is given by the following formula:

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

where the vertex is (h, k). Thus, in our equation k = -1 and h = 5, and the vertex

is (5, -1).

Additionally, the focus is given by (h+p, k). In our case:

$p=\frac{6}{4}=\frac{3}{2}$

Then, the focus is:

$(5+\frac{3}{2},-1)$

Simplifying:

$(\frac{13}{2},-1)$

The directrix is x = h - p:

$x=5-\frac{3}{2}=\frac{7}{2}$

Finally, the axis of symmetry is y = -1.