Question

determine the following for the parabola given by the equation

( y ^ { 2 } + 4 y - 8 x - 20 = 0 )

find the focus.

( ( x , y ) )

find an equation for the directrix.

directrix

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

endpoints

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

Complete the square for the \(y\) - terms.

$$
LATEXBLOCK0
$$

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

Step2: Find the focus

The focus of a parabola \((y - k)^{2}=4p(x - h)\) that opens to the right is \((h + p,k)\).
Substitute \(h=-3,k=-2,p = 2\) into the formula: \((-3+2,-2)=(-1,-2)\)

Step3: Find the directrix

The directrix of a parabola \((y - k)^{2}=4p(x - h)\) that opens to the right is \(x=h - p\).
Substitute \(h=-3,p = 2\) into the formula: \(x=-3-2=-5\)

Step4: Find the endpoints of the latus rectum

The endpoints of the latus rectum of a parabola \((y - k)^{2}=4p(x - h)\) are \((h + p,k\pm2p)\)
Substitute \(h=-3,k=-2,p = 2\) into the formula:
For \(y=-2 + 2p\), \(y=-2+4 = 2\) and \(x=-1\); for \(y=-2-2p\), \(y=-2 - 4=-6\) and \(x=-1\)
The endpoints are \((-1,2),(-1,-6)\)

Answer:

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