Chapter 5, Problem 1RE

To determine

The vertex, focus, and the directrix of the parabola $y^2=-16x$ and sketch the graph.

Answer to Problem 1RE

The vertex, focus, and the directrix of the parabola $y^2=-16x$ are $\underset{\_}{V\left(0,0\right),F\left(-4,0\right)\text{ and }x=4}$ respectively.

Explanation of Solution

Formula used:

Equation of a parabola: Vertex at $\left(0,0\right)$; Focus on an axis $a>0$.

Vertex	Focus	Directrix	Equation	Axis of symmetry	Opens
$\left(0,0\right)$	$\left(a,0\right)$	$x=-a$	$y^2=4ax$	x-axis	Right
$\left(0,0\right)$	$\left(-a,0\right)$	$x=a$	$y^2=-4ax$	x-axis	Left
$\left(0,0\right)$	$\left(0,a\right)$	$y=-a$	$x^2=4ay$	y-axis	Up
$\left(0,0\right)$	$\left(0,-a\right)$	$y=a$	$x^2=-4ay$	y-axis	Down

Calculation:

The equation of the parabola is $y^2=-16x$.

Notice that the equation is of the form, $y^2=-4ax$ whose vertex is $V\left(0,0\right)$, the focus is $\left(-a,0\right)$ and the directrix is $x=a$.

Comparing the equation, $y^2=-4ax$ and $y^2=-16x$, $4a=16$.

Thus, $a=4$.

Therefore, the vertex is $\underset{\_}{V\left(0,0\right)}$, focus is $\underset{\_}{\left(-4,0\right)}$ and the directrix is $\underset{\_}{x=4}$. Hence, the parabola opens upward.

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To draw the graph of the parabola, find two points to the left and right of the parabola.

Substitute, $x=-4$ in $y^2=-16x$,

$\begin{array}{l}{y}^2=-16\left(-4\right)\\ y^2=64\\ y=\sqrt{64}\\ y=\pm 8\end{array}$

The points to the left and the right of the focus are, $\left(-4,8\right)\text{ and }\left(-4,-8\right)$.

Use the above information and draw the graph of the parabola $y^2=-16x$ as shown below in Figure 1.

From Figure 1 note that the parabola opens left.