Chapter 10.2, Problem 42AYU

In Problems 39-56, find the vertex, focus, and directrix of each parabola. Graph the equation by hand. Verify your graph using a graphing utility.

${\left(x+\text{ 4}\right)}^2\text{ = 16}\left(y\text{ + 2}\right)$

To determine

(a & b)

To find: Vertex, focus and directrix of the parabola given by, $\left(x+4\right)²=16\left(y+2\right)$.

Answer to Problem 42AYU

Vertex: $(-4,-2)$.

Focus: $(-4,2)$.

Directrix D: $y=-6$.

Explanation of Solution

Given:

$\left(x+4\right)²=16\left(y+2\right)$

Formula used:

Equation	Vertex	Focus	Directrix	Description
${\left(x-h\right)}^2=4a\left(y-k\right)$	$(h,k)$	$\left(h,a+k\right)$	$y=k-a$	Parabola, axis of symmetry is parallel to $y\text{-axis}$, opens up.

Calculation:

The equation $\left(x+4\right)²=16\left(y+2\right)$ is of the form ${\left(x-h\right)}^2=4a\left(y-k\right)$.

$k=-2;h=-4$

Hence, $4a=16$; $a=4$.

Therefore, vertex: $\left(h,k\right) =\left(-4,-2\right)$.

Focus: $\left(h,a+k\right)=\left(-4,2\right)$.

Directrix D: $y=k-a=-2-4=-6;y=-6$

Parabola, axis of symmetry is parallel to $y\text{-axis}$, and it opens up.

To determine

c.

To graph: The parabola for the equation $\left(x+4\right)²=16\left(y+2\right)$ by hand verify using a graphing tool.

Answer to Problem 42AYU

Explanation of Solution

Given:

$\left(x+4\right)²=16\left(y+2\right)$

Calculation:

The graph of the parabola $\left(x+4\right)²=16\left(y+2\right)$.

$\left(x+4\right)²=16\left(y+2\right)$

By using the table above, we obtain few points $\left(-2,-1.75\right),\left(-8,-1\right),\left(0,-1\right)\left(-8,-1\right)$ and $\left(-6,-1.75\right)$. We plot the graph of $\left(x+4\right)²=16\left(y+2\right)$ using these points and the graph opens to the up. The points are shown above in the graph.

Using graphing utility we verify the graph.

The Vertex $\text{V}\left(-4, -2\right)$, focus $\text{F}\left(-4, 2\right)$ is shown in the graph above. The directrix D: $y=-6$ is shown in the dotted line.