Chapter 10.2, Problem 33AYU

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 $\left(-3,4\right)$; directrix the line $y\text{ = 2}$

To determine

a & b

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

Focus at $\left(-3,4\right)$; directrix the line $y=2$

Answer to Problem 33AYU

${\left(x+3\right)}^2=4\left(y-3\right)$ Latus rectum:$(-5,4)$ and $(-1,4)$.

Explanation of Solution

Given:

Focus at $\left(-3, 4\right)$; directrix the line $y=2$

Formula used:

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

Calculation:

The focus is at $F(-3,4)$ and the directrix the line $y=2$. Since $y=2$, axis of symmetry is parallel to $y\text{-axis}$. Focus at $(-3,4)$ and directrix the line $y=2$ implies,

The Vertex is at $V\left(-3,3\right)$ ($y$ coordinate of $V$ is 3 because the distance between vertex and focus is $\left(4-2\right)/2=\frac{2}{2}=1)$ and hence $a=1$

Focus lies above the Vertex. Hence the parabola opens up.

Hence the equation of parabola is,

${\left(x-\left(-3\right)\right)}^2=4a\left(y-3\right)$

${\left(x+3\right)}^2=4a\left(y-3\right)$

By substituting $a=1$

${\left(x+3\right)}^2=4a\left(y-3\right)$

The equation of the parabola is,

${\left(x+3\right)}^2=4\left(y-3\right)$

Directrix D: $y=2$

Substituting $y=4$ we get,

${\left(x+3\right)}^2=4*1$

${\left(x+3\right)}^2=4$

$x+3=\pm 2$

$x=-5, x=-1$

Hence $\left(-5, 4\right)$ and $\left(-1, 4\right)$ defines the latus rectum.

To determine

c.

To graph: The parabola for the equation ${\left(x+3\right)}^2=4\left(y-3\right)$

Answer to Problem 33AYU

Explanation of Solution

Given:

Focus at $(-3,4)$; directrix the line $y=2$

Calculation:

The graph of the parabola ${\left(x+3\right)}^2=4\left(y-3\right)$

Using the given information Focus at $\left(-3, 4\right)$; directrix the line $y=2$, we find Vertex $V\left(-3, 3\right)$ and we have formed the equation and plotted the graph for ${\left(x+3\right)}^2=4\left(y-3\right)$

The focus $F\left(-3, 4\right)$ and the vertex $V\left(-3, 3\right)$ lie on the vertical line $x=-3$ as shown. The equation for directrix D: $y=2$ is shown as dotted line. The parabola opens up.

The points $\left(-5, 4\right)$ and $\left(-1, 4\right)$ defines the latus rectum.