Question

question 2 (2 points)
(02.08)
derive the equation of the parabola with a focus at (-5, 5) and a directrix of y = -1.
a
$f(x)=-\frac{1}{12}(x - 5)^2+2$
b
$f(x)=\frac{1}{12}(x - 5)^2+2$
c
$f(x)=-\frac{1}{12}(x + 5)^2+2$
d
$f(x)=\frac{1}{12}(x + 5)^2+2$

Explanation:

Step1: Find the vertex

The vertex of a parabola is the mid - point between the focus and the point on the directrix directly below (or above) the focus. The x - coordinate of the vertex is the same as the x - coordinate of the focus, $x=-5$. The y - coordinate of the vertex $h$ is $\frac{5+( - 1)}{2}=\frac{5 - 1}{2}=2$. So the vertex is $(-5,2)$.

Step2: Determine the value of $a$

The distance $p$ between the vertex and the focus (or the vertex and the directrix) is $|5 - 2| = 3$. For a parabola of the form $y=a(x - h)^2+k$, the relationship between $a$ and $p$ is $a=\frac{1}{4p}$. Since $p = 3$, then $a=\frac{1}{12}$.

Step3: Write the equation of the parabola

The standard form of a parabola with vertex $(h,k)$ is $y=a(x - h)^2+k$. Here, $h=-5,k = 2,a=\frac{1}{12}$. Substituting these values, we get $y=\frac{1}{12}(x + 5)^2+2$, or $f(x)=\frac{1}{12}(x + 5)^2+2$.

Answer:

D. $f(x)=\frac{1}{12}(x + 5)^2+2$