Chapter 9.3, Problem 17WE

To determine

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

Answer to Problem 17WE

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

Explanation of Solution

Given information: The equation of the parabola is $y=\frac{1}{6}{x}^2$ .

Formula used: $y-k=\frac{1}{4c}{\left(x-h\right)}^2;c\ne 0$

Calculation: Since the vertex $\left(h,k\right)$ , axis $x=h$ , focus $\left(h,k+c\right)$ and directrix $y=k-c$ of the parabola $y-k=\frac{1}{4c}{\left(x-h\right)}^2;c\ne 0$ .

Hence, rewrite the given equation in to general form of equation,

  $y-0=\frac{1}{6}{\left(x-0\right)}^2$

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

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

Now,

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

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

The axis is $x=0$

• The directrix is,

  $\begin{array}{l}\Rightarrow y=k-c\\ =0-\frac{3}{2}\\ =\frac{-3}{2}\end{array}$

Now, graph the given parabola,

  

Interpretation:

The blue curve is the graph of $\left(y=\frac{1}{6}{x}^2\right)$.

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

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

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