Chapter 10.2, Problem 31AYU

In Problems 21-38, find the equation of the parabola described. Find the two points that define the latus rectum, and graph the equation by hand.

Vertex at $\left(-1,-2\right)$; focus at $\left(0,-2\right)$

To determine

The equation of the parabola described and two points that define the latus rectum for the following description.

Vertex at \left(-1,-2\right); focus at \left(0,-2\right).

Answer to Problem 31AYU

${\left(y+2\right)}^2=4\left(x+1\right)$, Latus rectum: $(0,-4)$ and $(0,0)$.

Explanation:

Given:

Vertex at \left(-1,-2\right); focus at \left(0,-2\right).

Formula used:

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

Calculation:

The vertex is at V\left(-1,-2\right) and the focus is at F\left(0,-2\right). Both lies on the line $x=-2$. Since focus F\left(0,-2\right) lies right to the vertex \left(-1,-2\right), the parabola opens right.

The length $a=\left|a\right|=\left|0-\left(-1\right)\right|=1$.

The equation for the parabola which opens right is given by,

$\left(y-\left(-2\right)\right)²=4a\left(x-\left(-1\right)\right)$

${\left(y+2\right)}^2=4a\left(x+1\right)$

By substituting $a=1$

${\left(y+2\right)}^2=4*1\left(x+1\right)$

${\left(y+2\right)}^2=4\left(x+1\right)$

The equation of the parabola is,

${\left(y+2\right)}^2=4\left(x+1\right)$

Directrix D: $x=h-a=-1-1=-2;x=-2$

Substituting $x=0$ we get,

${\left(y+2\right)}^2=4*1*\left(0+1\right)$

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

$y+2=\pm 2$

$y=-4,y=0$

Hence $\left(0,-4\right)$ and $\left(0,0\right)$ defines the latus rectum.

Explanation of Solution

Given:

Vertex at \left(-1,-2\right); focus at \left(0,-2\right).

Formula used:

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

Calculation:

The vertex is at V\left(-1,-2\right) and the focus is at F\left(0,-2\right). Both lies on the line $x=-2$. Since focus F\left(0,-2\right) lies right to the vertex \left(-1,-2\right), the parabola opens right.

The length $a=\left|a\right|=\left|0-\left(-1\right)\right|=1$.

The equation for the parabola which opens right is given by,

$\left(y-\left(-2\right)\right)²=4a\left(x-\left(-1\right)\right)$

${\left(y+2\right)}^2=4a\left(x+1\right)$

By substituting $a=1$

${\left(y+2\right)}^2=4*1\left(x+1\right)$

${\left(y+2\right)}^2=4\left(x+1\right)$

The equation of the parabola is,

${\left(y+2\right)}^2=4\left(x+1\right)$

Directrix D: $x=h-a=-1-1=-2;x=-2$

Substituting $x=0$ we get,

${\left(y+2\right)}^2=4*1*\left(0+1\right)$

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

$y+2=\pm 2$

$y=-4,y=0$

Hence $\left(0,-4\right)$ and $\left(0,0\right)$ defines the latus rectum.

To determine

c.

To graph: The parabola for the equation ${\left(y+2\right)}^2=4\left(x+1\right)$.

Answer to Problem 31AYU

Explanation of Solution

Given:

Vertex at \left(-1,-2\right); focus at \left(0,-2\right).

Calculation:

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

Using the given information vertex at V\left(-1,-2\right) and the focus at F\left(0,-2\right) we find $a=1$ and we have formed the equation and plotted the graph for ${\left(y+2\right)}^2=4\left(x+1\right)$.

The focus F\left(0,-2\right) and the vertex V\left(-1,-2\right) lie on the horizontal line $y=-2$ as shown. The equation for directrix D: $x=-2$ is shown as dotted line. The parabola opens right. The points \left(0,0\right) and \left(0,-4\right) defines the latus rectum.