Chapter 11, Problem 1RCC

1. (a) Give the geometric definition of a parabola.
2. (b) Give the equation of a parabola with vertex at the origin and with vertical axis. Where is the focus? What is the directrix?
3. (c) Graph the equation x² = 8y. Indicate the focus on the graph.

To determine

To state: The geometric definition of parabola.

Explanation of Solution

The geometric definition of parabola is stated as follows,

The set of points on a plane which are equidistance from a fixed point and a fixed line is called a parabola. Where the fixed point is called the focus and the fixed line is called the directrix.

To determine

To give: The equation of parabola with vertex at the origin and with vertical axis and find the focus and the directrix.

Answer to Problem 1RCC

The equation of the parabola with vertex at the origin and with vertical axis is $x^2=4py$.

When $p>0$, the focus $F\left(0,p\right)$ of the parabola $x^2=4py$ is at a distance of $p$ from the vertex $V(0,0)$ on the positivey-axis.

When $p<0$, the focus $F\left(0,-p\right)$ of the parabola $x^2=4py$ is at a distance of $p$ from the vertex $V(0,0)$ on the negativey-axis.

When $p>0$, the directrix of the parabola $x^2=4py$ is $y=-p$.

When $p<0$, the directrix of the parabola $x^2=4py$ is $y=p$.

Explanation of Solution

Definition used:

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

By the definition stated above the parabola $x^2=4py$ has the vertex at the origin $\left(0,0\right)$ and the focus is $F\left(0,p\right)$. It is observed that the focus $F\left(0,p\right)$ lies on y-axis, that is vertical axis. Since the focus always on the axis, the axis of the parabola $x^2=4py$ is a vertical axis.

In case of $p>0$, that is, $p$ is positive value then by the definition stated above the vertex and the focus of the parabola $x^2=4py$ are $V\left(0,0\right)$ and $F\left(0,p\right)$. It is observed that the focus $F(0,p)$ is at a distance of $p$ from the vertex $V\left(0,0\right)$ on the positive y-axis. Since the focus and directrix of the parabola $x^2=4py$ are at equal distance from the vertex $V(0,0)$ in the opposite directions, the equation of directrix is $y=-p$.

In case of $p<0$, that is, $p$ is negative value then by the definition stated above the vertex and the focus of the parabola $x^2=4py$ are $V(0,0)$ and $F(0,-p)$. It is observed that the focus $F(0,-p)$ is at a distance of $p$ from the vertex $V(0,0)$ on the negative y-axis. Since the focus and directrix of the parabola $x^2=4py$ are at equal distance from the vertex $V(0,0)$ in the opposite directions, the equation of directrix is $y=p$.

To determine

To graph: The parabola $x^2=8y$ and indicate its focus on the graph.

Answer to Problem 1RCC

The focus and directrix of the parabola $x^2=8y$ are $F(0,2)$ and $y=-2$ respectively.

The graph of the parabola $x^2=8y$ together with its focus $F\left(0,2\right)$ is shown below in Figure 1.

From Figure 1, it is observed that the focus is at $\left(0,2\right)$ and directrix is at the line $y=-2$.

Explanation of Solution

Compare the equation $x^2=8y$ with the parabola $x^2=4py$, it is clear that,

$\begin{array}{l}4p=8\\ p=2\end{array}$

Therefore, by the definition stated above, focus is $F\left(0,2\right)$, and the directrix is $y=-2$.

Thus, the focus and directrix of the parabola $x^2=8y$ are $F(0,2)$ and $y=-2$ respectively.

The graph of the parabola $x^2=8y$ with the vertex $V(0,0)$, focus $F\left(0,2\right)$ and directrix $y=-2$ is shown below in Figure 1.

From the Figure 1, it is observed that the focus is at $F\left(0,2\right)$ is indicated $2$ units above the vertex $V\left(0,0\right)$ and the directrix is at the line $y=-2$.