Question

1. $y = \frac{1}{8}(x - 2)^2+2$

vertex: (2,2)
$c = 2$
focus=(2,4)
directrix: $y=-2$

Explanation:

Step1: Identify the vertex - form of parabola

The equation $y = \frac{1}{8}(x - 2)^2+2$ is in the vertex - form $y=a(x - h)^2 + k$, where $(h,k)$ is the vertex. Here $h = 2$ and $k = 2$, so the vertex is $(2,2)$.

Step2: Find the value of $a$ and $c$

For a parabola $y=a(x - h)^2 + k$, the relationship between $a$ and the distance from the vertex to the focus $c$ is $a=\frac{1}{4c}$. Given $a=\frac{1}{8}$, we solve $\frac{1}{8}=\frac{1}{4c}$ for $c$. Cross - multiplying gives $4c = 8$, so $c = 2$.

Step3: Determine the focus

Since the parabola opens upwards (because $a=\frac{1}{8}>0$) and the vertex is $(h,k)=(2,2)$ and $c = 2$, the focus is $(h,k + c)$. Substituting the values, we get the focus $(2,2 + 2)=(2,4)$.

Step4: Find the directrix

For a parabola opening upwards with vertex $(h,k)$ and $c$ as the distance from the vertex to the focus, the equation of the directrix is $y=k - c$. Substituting $k = 2$ and $c = 2$, we get $y=2-2=0$.

Answer:

Vertex: $(2,2)$
$c = 2$
Focus: $(2,4)$
Directrix: $y = 0$