Chapter 10.2, Problem 25AYU

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 $y\text{ = }-\frac{1}{2}$; 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 $y\text{ = }-\frac{1}{2}$.

Answer to Problem 25AYU

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

Explanation of Solution

Given:

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

Formula used:

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

Calculation:

Given that the vertex is at $V\left(0,0\right)$ and the directrix is the line $y=-\frac{1}{2}$. The focus is $F\left(0,a\right)$ which is $F\left(0,\frac{1}{2}\right)$. Both vertex $V\left(0,0\right)$ and $F\left(0,\frac{1}{2}\right)$ lies on the vertical line $x=\text{ 0}$ (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 $F\left(0,\frac{1}{2}\right)$ lies up from the vertex, the parabola opens up.

$x^2=4ay$

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

Therefore the equation is,

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

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

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

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

We get,

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

$x=\pm 1$

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

To determine

c.

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

Answer to Problem 25AYU

Explanation of Solution

Given:

Focus at $\left(0,\frac{1}{2}\right)$; vertex at $\left(0,\text{ 0}\right)$.

Calculation:

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

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

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