Chapter 9.1, Problem 63E

To determine

Tofind:Thevertex, focus and directrix of the parabola $y^2=-6x$ and draw its graph. Verify the result graphically.

Answer to Problem 63E

The vertex, focus and directrix of the parabola $y^2=-6x$ are $\left(0,0\right)$ , $\left(-\frac{3}{2},0\right)$ , and $x=\frac{3}{2}$ respectively. The graph of the parabola is shown in figure (1).

Explanation of Solution

Given information:

Theequation of the parabola is $y^2=-6x$ .

Calculation:

Rearrange the equation of the parabola in the standard form.

  $\begin{array}{c}{y}^2=-6x\\ y^2=4\cdot \left(-\frac{6}{4}\right)x\\ y^2=4\cdot \left(-\frac{3}{2}\right)x\end{array}$

Write the expression for the standard form of the parabola when it had a vertical axis.

  ${\left(y-k\right)}^2=4p\left(x-h\right)$

Where,

• $p<0$ .

Compare the given equation of the parabola with the standard form of the parabola.

  $\begin{array}{c}h=0\\ p=-\frac{3}{2}\\ k=0\end{array}$

Determine the vertex.

  $\left(h,k\right)=\left(0,0\right)$

Calculate the focus.

  $\begin{array}{c}\left(h+p,k\right)=\left(0-\frac{3}{2},0\right)\\ =\left(-\frac{3}{2},0\right)\end{array}$

Calculate the directrix.

  $\begin{array}{c}x=k-p\\ x=0-\left(-\frac{3}{2}\right)\\ x=\frac{3}{2}\end{array}$

Draw the graph of the parabola.

  

Figure (1)

Verify the result using the graphical utility:

Draw the graph for the given parabola.

  

Therefore, the vertex, focus and directrix of the parabola $y^2=-6x$ are $\left(0,0\right)$ , $\left(-\frac{3}{2},0\right)$ , and $x=\frac{3}{2}$ respectively. The graph of the parabola is shown in figure (1).