Chapter 10.2, Problem 35AYU

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.

Focus at $\left(-3,-2\right)$; directrix the line $x\text{ = 1}$

To determine

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

Focus at $\left(-3,-2\right)$; directrix the line $x=1$

Answer to Problem 35AYU

${\left(y+2\right)}^2=-8\left(x+1\right)$ Latus rectum:$(-3,2)$ and $(-3,-6)$

Explanation of Solution

Given:

Focus at $\left(-3, -2\right)$; directrix the line $x=1$

Formula used:

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

Calculation:

The focus is at $F(-3,-2)$ and the directrix the line $x=1$. Since $x=1$, axis of symmetry is parallel to $x\text{-axis}$. Focus at $(-3,-2)$ and directrix the line $x=1$ implies,

The Vertex is at $V\left(-1,-2\right)$ and $a=2$($x$ coordinate of $V$ is $-1$ because the distance between directrix and focus is $1-\left(-3\right)=4=2a$)

Focus lies left of the Vertex. Hence the parabola opens left.

Hence the equation of parabola is

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

${\left(y-\left(-2\right)\right)}^2=-4a\left(x-\left(-1\right)\right)$

By substituting $a=2$

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

The equation of the parabola is

${\left(y+2\right)}^2=-8\left(x+1\right)$

Substituting $x=-3$ we get,

${\left(y+2\right)}^2=-8\left(-3+1\right)$

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

$y+2=\pm 4$

$y=2, y=-6$

Hence $\left(-3, 2\right)$ and $\left(-3, -6\right)$ defines the latus rectum.

To determine

To graph: The parabola for the equation ${\left(y+2\right)}^2=-8\left(x+1\right)$.

Answer to Problem 35AYU

Explanation of Solution

Given:

Focus at $(-3,-2)$; directrix the line $x=1$.

Calculation:

The graph of the parabola ${\left(y+2\right)}^2=-8\left(x+1\right)$.

Using the given information Focus at $\left(-3, -2\right)$; directrix the line $x=1$, we find Vertex $V\left(-1, -2\right)$ and we have formed the equation and plotted the graph for${\left(y+2\right)}^2=-8\left(x+1\right)$

The focus $F\left(-3, -2\right)$ and the vertex $V\left(-1, -2\right)$ lie on the horizontal line $y=-2$ as shown. The equation for directrix D: $x=1$ is shown as dotted line. The parabola opens left. The points $\left(-3, -2\right)$ and $\left(-3, -6\right)$ defines the latus rectum.