Question

determine the following for the parabola given by the equation
$y^2 = 24x$

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:

Part 1: Find the focus

Step1: Recall the standard form of a parabola

The standard form of a parabola that opens to the right (or left) is \( y^2 = 4px \), where the focus is at \( (p, 0) \) and the directrix is \( x=-p \).
Given the equation \( y^2 = 24x \), we compare it with \( y^2 = 4px \). So, \( 4p = 24 \).

Step2: Solve for \( p \)

To find \( p \), we divide both sides of \( 4p = 24 \) by 4: \( p=\frac{24}{4}=6 \).
Since the parabola is of the form \( y^2 = 4px \) (opening to the right), the focus is at \( (p, 0)=(6, 0) \).

Step1: Use the value of \( p \) from before

For the standard form \( y^2 = 4px \), the directrix is \( x=-p \). We found \( p = 6 \) in the previous part.

Step2: Substitute \( p \) into the directrix formula

Substituting \( p = 6 \) into \( x=-p \), we get \( x=-6 \).

Step1: Recall the definition of the latus rectum

The latus rectum of a parabola \( y^2 = 4px \) is a line segment perpendicular to the axis of the parabola (the x - axis here) passing through the focus. The length of the latus rectum is \( |4p| \), and its endpoints have the same x - coordinate as the focus (since it passes through the focus) and their y - coordinates are \( \pm 2p \) (because the length of the latus rectum is \( 4p \), so from \( y=-2p \) to \( y = 2p \) when \( x = p \)).
We know \( p = 6 \), so the x - coordinate of the endpoints of the latus rectum is \( x = p=6 \), and the y - coordinates are \( y=\pm 2p=\pm 12 \).

Step2: Write the endpoints

So the endpoints of the latus rectum are \( (6, 12) \) and \( (6, - 12) \).

Answer:

\( (6, 0) \)

Part 2: Find the equation for the directrix