Find the vertex focus and directrix of Y-4=1/16(x-7)^2

Y - 4 = \frac{1}{16}(x - 7)^2

[SOLUTION] \text{Vertex: }(7,4),\text{ Focus: }(7,8),\text{ Directrix: }y=0

[STEP-BY-STEP]

Y - 4 = \frac{1}{16}(x - 7)^2

Compare with the standard form of a parabola opening upwards:

y = a(x - h)^2 + k

where:

(h, k) \text{ is the vertex, and } a \text{ determines the focal length.}

From the given equation, we identify:

h = 7, k = 4, a = \frac{1}{16}

The vertex is:

(h, k) = (7, 4)

The focal length \( \frac{1}{4a} \):

\frac{1}{4a} = \frac{1}{4 \cdot \frac{1}{16}} = 4

Focus:

(h,k+\frac{1}{4a})=(7,4+4)=(7,8)

Directrix:

y=k-\frac{1}{4a}=4-4=0

Answer:

\text{Vertex: } (7, 4), \text{ Focus: } (7, 8), \text{ Directrix: } y = 0