Question

consider the parabola
-2x^{2}+16x - 4y - 16 = 0
find the vertex. enter your answer as an ordered pair.

find the focus. enter your answer as an ordered pair.

find an equation for the directrix.

graph the parabola using the focus and directrix.

Explanation:

Step1: Rewrite the equation in standard form

Start with \(-2x^{2}+16x - 4y-16 = 0\).
First, isolate \(y\):

$$
LATEXBLOCK0
$$

For a parabola \(y = ax^{2}+bx + c\), the \(x\) - coordinate of the vertex is \(x=-\frac{b}{2a}\). Here \(a =-\frac{1}{2}\), \(b = 4\).

$$x=-\frac{4}{2\times(-\frac{1}{2})}=4$$

Substitute \(x = 4\) into \(y=-\frac{1}{2}x^{2}+4x - 4\):

$$y=-\frac{1}{2}(4)^{2}+4\times4-4=-\frac{1}{2}\times16 + 16-4=-8 + 16-4 = 4$$

The vertex form of a parabola is \(y=a(x - h)^{2}+k\), where \((h,k)\) is the vertex. We can also complete the square:

$$
LATEXBLOCK1
$$

The vertex is \((4,4)\).

Step2: Find the focus

The standard form of a parabola is \((x - h)^{2}=4p(y - k)\). Rewrite our equation \(y=-\frac{1}{2}(x - 4)^{2}+4\) as \((x - 4)^{2}=-2(y - 4)\).
Comparing with \((x - h)^{2}=4p(y - k)\), we have \(4p=-2\), so \(p=-\frac{1}{2}\).
The focus of the parabola \((x - h)^{2}=4p(y - k)\) is \((h,k + p)\). Here \(h = 4\), \(k = 4\), \(p=-\frac{1}{2}\)
The focus is \((4,4-\frac{1}{2})=(4,\frac{7}{2})\)

Step3: Find the directrix

The equation of the directrix for the parabola \((x - h)^{2}=4p(y - k)\) is \(y=k - p\).
Since \(k = 4\), \(p=-\frac{1}{2}\), then \(y=4-(-\frac{1}{2})=\frac{9}{2}\)

Answer:

• Vertex: \((4,4)\)
• Focus: \((4,\frac{7}{2})\)
• Directrix: \(y=\frac{9}{2}\)