Chapter 10.2, Problem 26AYU

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.

Directrix the line $x\text{ = }-\frac{1}{2}$; vertex at $\left(0,\text{ 0}\right)$

To determine

(a & b)

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

Vertex at $\left(0,\text{ 0}\right)$; directrix the line $x\text{ = }-\frac{1}{2}$.

Answer to Problem 26AYU

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

Explanation of Solution

Given:

vertex at $\left(0,\text{ 0}\right)$; directrix the line $x\text{ = }-\frac{1}{2}$.

Formula used:

Vertex	Focus	Directrix	Equation	Description
$\left(0,\text{ 0}\right)$	$\left(a,\text{ 0}\right)$	$x=-a$	$y^2=4ax$	Parabola, axis of symmetry is $x\text{-axis}$, opens right

Calculation:

Given that the vertex is at $V\left(0,0\right)$ and the directrix is the line $x\text{ = }-\frac{1}{2}$. The focus is $F\left(a,\text{ 0}\right)$ which is $F\left(\frac{1}{2},\text{ 0}\right)$. Both vertex $V\left(0,0\right)$ and $F\left(\frac{1}{2},\text{ 0}\right)$ lies on the horizontal line $y=\text{ 0}$ (The axis of symmetry). The distance $a$ from the vertex $V\left(0,0\right)$ to the focus $F\left(0,\frac{1}{2}\right)$ is $a=\frac{1}{2}$.

Also, because the focus $F\left(\frac{1}{2},\text{ 0}\right)$ lies right of he vertex, the parabola opens right.

$y^2=4ax$

$a=\frac{1}{2}$

Therefore the equation is,

$y^2=4*\frac{1}{2}*x$

$y^2=\text{ 2}x$

The equation of the parabola is $y^2=\text{ 2}x$.

To find the points that define the latus rectum, Let $x\text{ = }\frac{1}{2}$ in the above equation.

We get,

$y^2=\text{ 2 }*\frac{1}{2}$

$y=\pm 1$

Hence $\left(\frac{1}{2},\text{ 1}\right)$ and $\left(\frac{1}{2},-1\right)$ defines the latus rectum.

To determine

c.

To graph: The parabola for the equation $y^2=\text{ 2}x$.

Answer to Problem 26AYU

Explanation of Solution

Given:

Vertex at $\left(0,\text{ 0}\right)$; directrix the line $x\text{ = }-\frac{1}{2}$.

Calculation:

Using the given information D: $x\text{ = }-\frac{1}{2}$ and $V\left(0,0\right)$, we find $F\left(\frac{1}{2},\text{ 0}\right)$ and we have formed the equation and plotted the graph for $y^2=\text{ 2}x$. The focus $F\left(\frac{1}{2},\text{ 0}\right)$ and the vertex $V\left(0,0\right)$ lie on the horizontal line $y=\text{ 0}$ as shown.

The equation for directrix D: $x\text{ = }-\frac{1}{2}$ is shown as dotted line. The parabola opens right. The points $\left(\frac{1}{2},\text{ 1}\right)$ and $\left(\frac{1}{2},-1\right)$ define the latus rectum.