Question

determine the following for the parabola given by the equation

find the focus.

find an equation for the directrix.

find the endpoints of the latus rectum. enter your answer as ordered pairs separated by a comma.

graph the parabola using the focus and the directrix. to move the focus and directrix, drag the black points.

Explanation:

Step1: Rewrite the equation in standard form

The given equation is \(x^{2}+4x + 8y+28 = 0\).
Complete the square for the \(x\) - terms:

$$
LATEXBLOCK0
$$

The standard form of a parabola is \((x - h)^{2}=4p(y - k)\), where \((h,k)\) is the vertex.
Comparing \((x + 2)^{2}=-8(y + 3)\) with \((x - h)^{2}=4p(y - k)\), we have \(h=-2,k=-3\), and \(4p=-8\), so \(p=-2\).

Step2: Find the focus

The formula for the focus of a parabola \((x - h)^{2}=4p(y - k)\) is \((h,k + p)\).
Substituting \(h=-2,k=-3,p=-2\) into the formula:
\(k + p=-3+(-2)=-5\)
The focus is \((-2,-5)\)

Step3: Find the directrix

The formula for the directrix of a parabola \((x - h)^{2}=4p(y - k)\) is \(y=k - p\).
Substituting \(k=-3,p=-2\) into the formula:
\(y=-3-(-2)=-1\)
The equation of the directrix is \(y=-1\)

Step4: Find the endpoints of the latus rectum

The length of the latus rectum is \(|4p| = 8\).
For a parabola \((x - h)^{2}=4p(y - k)\), the endpoints of the latus rectum are \((h\pm2p,k + p)\)
Since \(h=-2,k=-3,p=-2\), \(2p=-4\)
The \(x\) - coordinates: \(x=-2\pm(-4)\)
When \(x=-2-4=-6\), \(y=-5\)
When \(x=-2 + 4 = 2\), \(y=-5\)
The endpoints of the latus rectum are \((-6,-5),(2,-5)\)

Answer:

• Focus: \((-2,-5)\)
• Directrix: \(y = - 1\)
• Endpoints of the latus rectum: \((-6,-5),(2,-5)\)