Chapter 8, Problem 3RE

Solution

To find: The vertex, focus, directrix and focal width of the parabola ${\left(x+2\right)}^2=-4\left(y-1\right)$ and sketch the graph.

The vertex is $\left(-2,1\right)$ , the focus is $\left(-2,0\right)$ , the directrix is $y=2$ , and the focal width is $4$ .

Given information:

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

Formula used:

For the graph of ${\left(x-h\right)}^2=4p\left(y-k\right)$ :

Vertex is $\left(h,k\right)$ .

Focus is $\left(h,k+p\right)$

Directrix is $y=k-p$

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

Calculation:

Compare the equation of the parabola ${\left(x+2\right)}^2=-4\left(y-1\right)$ with the standard form of equation of parabola, ${\left(x-h\right)}^2=4p\left(y-k\right)$ .

  $\begin{array}{c}4p=-4\\ p=\frac{-4}{4}\\ p=-1\\ \left|p\right|=\left|-1\right|\\ =1\end{array}$

Substitute $-2$ for $h$ and $1$ for $k$ in the vertex formula.

Vertex is $\left(-2,1\right)$

Substitute $-2$ for $h$ and $1$ for $k$ and $-1$ for $p$ in the formula to find Focus.

  $\begin{array}{l}\left(-2,1-1\right)\\ =\left(-2,0\right)\\ =\left(-2,0\right)\end{array}$ .

Substitute $1$ for $k$ and $-1$ for $p$ in the formula to find Directrix.

  $\begin{array}{c}y=1-\left(-1\right)\\ =1+1\\ =2\end{array}$

Focal width is $4$ .

Draw the graph as follows:

  

Hence, the vertex is $\left(-2,1\right)$ , the focus is $\left(-2,0\right)$ , the directrix is $y=2$ , and the focal width is $4$ .