Chapter 10.2, Problem 27AYU

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 $\left(0,\text{ 0}\right)$; axis of symmetry the $y\text{-axis}$; containing the point $\left(2,\text{ 3}\right)$

To determine

(a & b)

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

Vertex at $\left(0,\text{ 0}\right)$; axis of symmetry the $y\text{-axis}$; containing the point $\left(2,3\right)$.

Answer to Problem 27AYU

$x^2=\frac{4}{3}y$, Latus rectum: $\left(\frac{2}{3},\frac{1}{3}\right)$ and $\left(-\frac{2}{3},\frac{1}{3}\right)$.

Explanation of Solution

Given:

Vertex at $\left(0,\text{ 0}\right)$; axis of symmetry the $y\text{-axis}$; containing the point $\left(2,3\right)$.

Formula used:

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

Calculation:

The vertex $V\left(0,0\right)$ and the axis of symmetry is vertical, hence the parabola opens up or down. But it is given that axis of symmetry contains $\left(2,3\right)$ and it is above $\left(0,\text{ 0}\right)$. Hence the parabola opens up. Hence the equation has the form $x^2=4ay$.

Substituting $\left(2,3\right)$ in the equation $x^2=4*a*y$ we get,

$4=4a\left(3\right)$

$a=\frac{1}{3}$

Hence , the focus is $\left(\text{0, }\frac{1}{3}\right)$.

The equation of the parabola is $x^2=\frac{4}{3}y$.

To find the points that define the latus rectum, Let $y\text{ = }\frac{1}{3}$ in the above equation.

We get,

$x^2=4*\frac{1}{3}*\frac{1}{3}$

$x=\pm \frac{2}{3}$

Hence $\left(\frac{2}{3},\frac{1}{3}\right)$ and $\left(-\frac{2}{3},\frac{1}{3}\right)$ defines the latus rectum.

To determine

c.

To graph: The parabola for the equation $x^2=\frac{4}{3}y$.

Answer to Problem 27AYU

Explanation of Solution

Given:

Vertex at $\left(0,\text{ 0}\right)$; axis of symmetry the $y\text{-axis}$; containing the point $\left(2,3\right)$.

Calculation:

The graph of the parabola $x^2=\frac{4}{3}y$.

Using the given information $V\left(0,0\right)$ and the point $\left(2,3\right)$ that passes through the axis symmetry the $y\text{-axis}$, we find $F\left(0,\frac{1}{3}\right)$ and we have formed the equation and plotted the graph for $x^2=\frac{4}{3}y$. The focus $F\left(0,\frac{1}{3}\right)$ and the vertex $V\left(0,0\right)$ lie on the vertical line $x\text{ = 0}$ as shown.

The equation for directrix D: $y\text{ = }-\frac{1}{3}$ is shown as dotted line. The parabola opens up. The points $\left(\frac{2}{3},\frac{1}{3}\right)$ and $\left(-\frac{2}{3},\frac{1}{3}\right)$ defines the latus rectum.