Chapter 11.1, Problem 4E

Label the focus, directrix, and vertex on the graphs given for the parabolas in Exercises 2 and 3.

1. (a) x² = 12y

1. (b) y² = 12x

To determine

To label: The focus, directrix and vertex on the graphs of the parabola $x^2=12y$.

Answer to Problem 4E

The focus, directrix and the vertex of the parabola $x^2=12y$ are labeled as shown below in Figure 1.

Explanation of Solution

Definition used:

 “The equation of the parabola vertex $V(0,0)$, focus $F(0,p)$ and directrix $y=-p$ is $x^2=4py$.”

Calculation:

The equation $x^2=12y$ is of the form $x^2=4py$, therefore

$\begin{array}{c}4p=12\\ p=3\end{array}$

By the definition of parabola stated above, vertex $V(0,0)$, focus $F(0,3)$ and the directrix is $y=-3$.

Use online graphing calculator and draw the graph of parabola $x^2=12y$ as shown below in Figure 1.

Observation:

The vertex $V(0,0)$ is the origin, hence it is marked at the intersection of coordinate axes on the Figure 1.

The focus $F(0,3)$ is on the positive y-axis at a distance of $3$ units from the vertex, hence it is marked on the y- axis on the Figure 1.

The directrix is at a distance of 3 units below the vertex $V(0,0)$, hence the line $y=-3$ is marked as directrix on the Figure 1.

To determine

To label: The focus, directrix and vertex on the graphs of the parabola $y^2=12x$.

Answer to Problem 4E

The focus, directrix and the vertex of the parabola is $y^2=12x$ are labeled as shown below in Figure 2.

Explanation of Solution

Definition used:

 “The equation of the parabola vertex $V(0,0)$, focus $F(p,0)$ and directrix $x=-p$ is $y^2=4px$”

Calculation:

The equation $y^2=12x$ is of the form $y^2=4px$, therefore

$\begin{array}{c}4p=12\\ p=3\end{array}$

By the definition of parabola stated above, $V(0,0)$, focus $F(3,0)$ and the directrix is $x=-3$.

Use online graphing calculator and draw the graph of parabola $y^2=12x$ as shown below in Figure 2.

Observation:

The vertex $V(0,0)$ is the origin, hence it is marked at the intersection of coordinate axes on the Figure 2.

The focus $F(3,0)$ is on the positive x-axis at a distance of 3 units from the vertex, hence it is marked on the x- axis on the Figure 2.

The directrix is at a distance of 3 units left of the vertex $V(0,0)$, hence the line $x=-3$ is marked as directrix on the Figure 2.