Chapter 8, Problem 1RE

Solution

To find: The vertex, focus, directrix and focal width of the parabola $y^2=12x$ and sketch the graph.

The vertex is $\left(0,0\right)$ , the focus is $\left(3,0\right)$ , the directrix is $x=-3$ , and the focal width is $12$ .

Given information:

Equation of the parabola $y^2=12x$ .

Formula used:

For the graph of $y^2=4px$ :

Vertex is $\left(0,0\right)$ .

Focus is $\left(p,0\right)$

Directrix is $x=-p$

Focal width is $\left|4p\right|$

Calculation:

Compare the equation of the parabola $y^2=12x$ with the standard form of equation of parabola, $y^2=4px$ .

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

Therefore, the vertex is $\left(0,0\right)$

Substitute $3$ for $p$ in the formula to find focus.

Focus is $\left(3,0\right)$ .

Substitute $3$ for $p$ in the formula to find directrix.

Directrix is $x=-3$ .

Substitute $3$ for $p$ in the formula to find focal width.

Focal width is $12$ .

Draw the graph as follows:

  

Figure 1

Hence, the vertex is $\left(0,0\right)$ , the focus is $\left(3,0\right)$ , the directrix is $x=-3$ , and the focal width is $12$ .