Chapter 10.2, Problem 51AYU

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.

$y^2+\text{ 2}y-x=0$

To determine

(a & b)

To find: Vertex, focus and directrix of the parabola given by, $y²+2y-x=0$.

Answer to Problem 51AYU

Vertex: $(-1,-1)$.

Focus: $\left(-\frac{3}{4},-1\right)$.

Directrix D: $x=-5/4$.

Explanation of Solution

Given:

$y²+2y-x=0$

Formula used:

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

Calculation:

The equation, $y²+2y-x=0$.

${\left(y+1\right)}^2=\left(x+1\right)$ is of the form ${\left(y-k\right)}^2=4a\left(x-h\right)$.

$h=-1;k=-1$

Hence, $4a=1;$ $a=1/4$.

Therefore, vertex: $\left(h,k\right) =\left(-1,-1\right)$.

Focus: $\left(h+a,k\right)=\left(-\frac{3}{4},- 1\right)$.

Directrix D: $x=h-a=-1-1/4=-5/4; x=-5/4$.

Parabola, axis of symmetry is parallel to $x\text{-axis}$, and it opens to the right.

To determine

c.

To graph: The parabola for the equation $y²+2y-x=0$ by hand and verify using a graphing tool.

Answer to Problem 51AYU

Explanation of Solution

Given:

$y²+2y-x=0$

Calculation:

The graph of the parabola $y²+2y-x=0$.

$y²+2y-x=0$

By using the table above, we obtain few points $\left(0,0\right),\left(0,-2\right)$ and $\left(-1,-1\right)$. We plot the graph of $y²+2y-x=0$ using these points and the graph opens to the right. The points are shown above in the graph.

Using graphing utility we verify the graph.

The Vertex $V\left(-1,-1\right)$, focus $F\left(-\frac{3}{4},-1\right)$ is shown in the graph above. The directrix D: $x=-\frac{5}{4}$ is shown in the dotted line.