Chapter 9.1, Problem 70E

To determine

To find: The vertex, focus and directrix of the parabola ${\left(x+\frac{1}{2}\right)}^2=4\left(y-1\right)$ and verify with the help of graphing utility.

Answer to Problem 70E

The vertex, focus and directrix of the parabola ${\left(x+\frac{1}{2}\right)}^2=4\left(y-1\right)$ are $\left(-\frac{1}{2},1\right)$ , $\left(-\frac{1}{2},2\right)$ , $y=0$ respectively. The graph of the parabola with vertex, focus and directrix is shown in Figure (1).

Explanation of Solution

Given information:

The equation of the parabola is ${\left(x+\frac{1}{2}\right)}^2=4\left(y-1\right)$ .

Calculation:

Compare the given equation of parabola with general vertical axis parabola equation ${\left(x-h\right)}^2=4p\left(y-k\right)$ . This gives the value of $h$ , $k$ , $p$ as $-\frac{1}{2}$ , $1$ , $1$ respectively.

The vertex of the parabola ${\left(x+\frac{1}{2}\right)}^2=4\left(y-1\right)$ is $\left(h,k\right)=\left(-\frac{1}{2},1\right)$ .

The focus of the horizontal axis parabola ${\left(x+\frac{1}{2}\right)}^2=4\left(y-1\right)$ is:

  $\begin{array}{c}\left(h,k+p\right)=\left(-\frac{1}{2},1+1\right)\\ =\left(-\frac{1}{2},2\right)\end{array}$

The directrix of the parabola ${\left(x+\frac{1}{2}\right)}^2=4\left(y-1\right)$ is:

  $\begin{array}{c}y=k-p\\ =1-1\\ =0\end{array}$

Therefore, The vertex, focus and directrix of the parabola ${\left(x+\frac{1}{2}\right)}^2=4\left(y-1\right)$ are $\left(-\frac{1}{2},1\right)$ , $\left(-\frac{1}{2},2\right)$ , $y=0$ respectively.

The graph of the parabola ${\left(x+\frac{1}{2}\right)}^2=4\left(y-1\right)$ with vertex, focus and directrix is shown below.

  

Figure (1)

Here we can verify the vertex, focus, directrix of the parabola ${\left(x+\frac{1}{2}\right)}^2=4\left(y-1\right)$ are $\left(-\frac{1}{2},1\right)$ , $\left(-\frac{1}{2},2\right)$ , $y=0$ respectively.