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: Identify the standard form of the parabola equation

The standard form of a parabola equation that opens up or down is \(x^{2}=4py\). Given \(x^{2}=-24y\), we can compare it with \(x^{2}=4py\). So, \(4p=-24\).

Step2: Solve for \(p\)

Divide both sides of the equation \(4p = - 24\) by \(4\). We get \(p=\frac{-24}{4}=-6\).

Step3: Find the focus

For a parabola of the form \(x^{2}=4py\), the focus is at the point \((0,p)\). Since \(p=-6\), the focus is \((0, - 6)\).

Step4: Find the directrix

The equation of the directrix for a parabola \(x^{2}=4py\) is \(y=-p\). Substituting \(p = - 6\), we get \(y = 6\).

Step5: Find the endpoints of the latus - rectum

The length of the latus - rectum is \(|4p|\). Here \(|4p| = 24\). For the parabola \(x^{2}=4py\), when \(y = p\), \(x^{2}=4p\times p\). Substituting \(p=-6\), we have \(x^{2}=4\times(-6)\times(-6)=144\), so \(x=\pm12\). The endpoints of the latus - rectum are \((-12,-6)\) and \((12,-6)\)

Answer:

• Focus: \((0,-6)\)
• Directrix: \(y = 6\)
• Endpoints of the latus - rectum: \((-12,-6),(12,-6)\)