Chapter 9.3, Problem 18WE

To determine

To calculate: The vertex, focus, directrix, and axis of symmetry.

Answer to Problem 18WE

The required vertex, focus, directrix and axis of symmetry of the parabola are $\left(0,0\right),\left(\frac{3}{2},0\right),x=-\frac{3}{2}$ and $\left(y=0\right)$ respectively.

Explanation of Solution

Given information: The equation of the parabola is $6x+y^2=0$ .

Formula used: $x-h=\frac{1}{4c}{\left(y-k\right)}^2$

Calculation: Rewrite the given equation in to general form of equation,

  $\begin{array}{l}6x+y^2=0\\ x-0=-\frac{1}{6}{\left(y-0\right)}^2\end{array}$

Now, compare $x-0=-\frac{1}{6}{\left(y-0\right)}^2$ to equation $x-h=\frac{1}{4c}{\left(y-k\right)}^2$ , we get,

  $\begin{array}{c}h=0\\ k=0\\ \begin{array}{l}y=k\\ \therefore y=0\end{array}\\ \begin{array}{l}\frac{1}{4c}=-\frac{1}{6}\\ -4c=-6\\ \therefore c=\frac{3}{2}\end{array}\end{array}$

Now,

The vertex is $\left(h,k\right)=\left(0,0\right)$

The focus is $\left(h+c,k\right)=\left(\frac{3}{2},0\right)$

The directrix is $x=h-c$

  $\begin{array}{l}=0-\frac{3}{2}\\ \therefore x=-\frac{3}{2}\end{array}$

The axis is $y=k=0$

Now, graph the given parabola,

  

Interpretation:

The blue curve is the graph of $6x+y^2=0$ .

The vertex of the parabola is $\left(0,0\right)$ .

The red vertical line is the directrix $\left(x=-\frac{3}{2}\right)$ .

The black horinzontal line is the axis of symmetry $\left(y=0\right)$ .