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consider the parabola $-2x^{2}+20x - 4y - 66 = 0$ find the vertex. ente…

Question

consider the parabola
$-2x^{2}+20x - 4y - 66 = 0$
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:

Response
Finding the Vertex

Step 1: Rewrite the equation in terms of y

Start with the given equation: \(-2x^2 + 20x - 4y - 66 = 0\)
We want to solve for \(y\), so first, move the \(x\)-terms and the constant to the other side:
\(-4y = 2x^2 - 20x + 66\)
Then divide both sides by \(-4\):
\(y = -\frac{1}{2}x^2 + 5x - \frac{33}{2}\)

Step 2: Complete the square for the \(x\)-terms

For a quadratic in the form \(y = ax^2 + bx + c\), the \(x\)-coordinate of the vertex is given by \(x = -\frac{b}{2a}\). Here, \(a = -\frac{1}{2}\) and \(b = 5\).
\(x = -\frac{5}{2 \times (-\frac{1}{2})} = -\frac{5}{-1} = 5\)
Now, substitute \(x = 5\) back into the equation to find the \(y\)-coordinate:
\(y = -\frac{1}{2}(5)^2 + 5(5) - \frac{33}{2}\)
\(y = -\frac{25}{2} + 25 - \frac{33}{2}\)
\(y = -\frac{25}{2} - \frac{33}{2} + 25\)
\(y = -\frac{58}{2} + 25\)
\(y = -29 + 25 = -4\)
So the vertex is \((5, -4)\)

Finding the Focus

For a parabola in the form \(y = ax^2 + bx + c\), the standard form is \((x - h)^2 = 4p(y - k)\), where \((h, k)\) is the vertex and \(p\) is the distance from the vertex to the focus (and also from the vertex to the directrix).
First, we can rewrite our equation \(y = -\frac{1}{2}x^2 + 5x - \frac{33}{2}\) in the form \((x - h)^2 = 4p(y - k)\).
Starting from \(y = -\frac{1}{2}x^2 + 5x - \frac{33}{2}\), multiply both sides by \(-2\) to make the coefficient of \(x^2\) equal to \(1\):
\(-2y = x^2 - 10x + 33\)
Now, complete the square on the right side. Take the coefficient of \(x\), which is \(-10\), divide by \(2\) to get \(-5\), and square it: \((-5)^2 = 25\)
Add and subtract \(25\) on the right side:
\(-2y = (x^2 - 10x + 25) + 33 - 25\)
\(-2y = (x - 5)^2 + 8\)
Now, solve for \((x - 5)^2\):
\((x - 5)^2 = -2(y + 4)\)
Comparing with \((x - h)^2 = 4p(y - k)\), we have \(h = 5\), \(k = -4\), and \(4p = -2\), so \(p = -\frac{2}{4} = -\frac{1}{2}\)
The focus is at \((h, k + p)\), so substituting the values:
\((5, -4 + (-\frac{1}{2})) = (5, -\frac{9}{2})\)

Finding the Directrix

The directrix is a horizontal line (since the parabola opens up or down) given by \(y = k - p\)
We know \(k = -4\) and \(p = -\frac{1}{2}\), so:
\(y = -4 - (-\frac{1}{2}) = -4 + \frac{1}{2} = -\frac{7}{2}\)

Vertex Answer:

\((5, -4)\)

Focus Answer:

\((5, -\frac{9}{2})\)

Directrix Answer:

\(y = -\frac{7}{2}\)

Answer:

The directrix is a horizontal line (since the parabola opens up or down) given by \(y = k - p\)
We know \(k = -4\) and \(p = -\frac{1}{2}\), so:
\(y = -4 - (-\frac{1}{2}) = -4 + \frac{1}{2} = -\frac{7}{2}\)

Vertex Answer:

\((5, -4)\)

Focus Answer:

\((5, -\frac{9}{2})\)

Directrix Answer:

\(y = -\frac{7}{2}\)