Question

question 5 (2 points)
(02 08)
derive the equation of the parabola with a focus at (4, -7) and a directrix of y = -15. put the equation in standard form.
(a) (f(x)=\frac{1}{16}x^{2}-8x + 11)
(b) (f(x)=\frac{1}{16}x^{2}-8x - 10)
(c) (f(x)=\frac{1}{16}x^{2}-\frac{1}{2}x + 11)
(d) (f(x)=\frac{1}{16}x^{2}-\frac{1}{2}x - 10)

Explanation:

Step1: Find the vertex

The vertex of a parabola is the mid - point between the focus $(4,-7)$ 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, $x = 4$. The $y$ - coordinate of the vertex is $\frac{-7+( - 15)}{2}=\frac{-7 - 15}{2}=-11$. So the vertex is $(h,k)=(4,-11)$.

Step2: Determine the value of $p$

The distance between the vertex and the focus (or the vertex and the directrix) is $p$. Since the focus is at $(4,-7)$ and the vertex is at $(4,-11)$, $p=-7-( - 11)=4$.

Step3: Write the standard - form equation

The standard - form equation of a parabola that opens up or down is $(x - h)^2 = 4p(y - k)$. Substituting $h = 4$, $k=-11$, and $p = 4$ into the equation, we get $(x - 4)^2=16(y + 11)$.
Expand the left - hand side: $x^{2}-8x + 16=16y+176$.
Solve for $y$:
\[

$$\begin{align*} 16y&=x^{2}-8x + 16-176\\ y&=\frac{1}{16}x^{2}-\frac{1}{2}x - 10 \end{align*}$$

\]

Answer:

D. $f(x)=\frac{1}{16}x^{2}-\frac{1}{2}x - 10$