Chapter 10.2, Problem 30AYU

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 $\left(4,-2\right)$; focus at $\left(6,-2\right)$

To determine

a & b

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

Vertex at \left(4,-2\right); focus at \left(6,-2\right).

Answer to Problem 30AYU

${\left(y+2\right)}^2=8\left(x-4\right)$, Latus rectum: $\underset{}{(6,-6)}$ and $(6,2)$.

Explanation of Solution

Given:

Vertex at \left(4,-2\right); focus at \left(6,-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 $x\text{-axis}$, opens right

Calculation:

The vertex is at V\left(4,-2\right) and the focus is at F\left(6,-2\right). Both lies on the line $y=-2$. Since focus F\left(6,-2\right) lies above the vertex \left(4,-2\right), hence, the parabola opens to the right. The length $a=\left|a\right|=\left|6-4\right|=2$.

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

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

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

By substituting $a=2$.

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

The equation of the parabola is,

${\left(y+2\right)}^2=8\left(x-4\right)$

Directrix D: $x=h-a=4-2=2;x=2$

Substituting $x=6$ we get,

${\left(y+2\right)}^2=8\left(6-4\right)$

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

$y+2=\pm 4$

$y=-6,y=2$

Hence $\left(6,-6\right)$ and $\left(6,2\right)$ defines the latus rectum.

To determine

c.

To graph: The parabola for the equation ${\left(y+2\right)}^2=8\left(x-4\right)$

Answer to Problem 30AYU

Explanation of Solution

Given:

Vertex at \left(4,-2\right); focus at \left(6,-2\right)

Calculation:

The graph of the parabola with the equation ${\left(y+2\right)}^2=8\left(x-4\right)$ is given by

Using the given information vertex at V\left(4,-2\right) and the focus at F\left(6,-2\right) we find $a=2$ and we have formed the equation and plotted the graph for ${\left(y+2\right)}^2=8\left(x-4\right)$.

The focus F\left(6,-2\right) and the vertex V\left(4,-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(6,2\right) and \left(6,-6\right) defines the latus rectum.