Chapter 10.2, Problem 36AYU

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

To determine

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

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

Answer to Problem 36AYU

${\left(x+4\right)}^2=12\left(y-1\right)$ Latus rectum:$(-10,4)$ and $(2,4)$

Explanation of Solution

Given:

Focus at $\left(-4, -4\right)$; 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 $x\text{-axis}$, opens up

Calculation:

The focus is at $F(-4,4)$ and the directrix the line $y=-2$. Since $y=-2$, axis of symmetry is parallel to $x\text{-axis}$. Focus at $(-4,4)$ and directrix the line $x=1$ implies,

The Vertex is at $V\left(-4,1\right)$ and $a=3$($y$ coordinate of $V$ is 1 because the distance between directrix and focus is $4-\left(-2\right)=6=2a$) and hence $a=3$

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

Hence the equation of parabola is

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

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

By substituting $a=3$

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

The equation of the parabola is

${\left(x+4\right)}^2=12\left(y-1\right)$

Substituting $y=4$ we get,

${\left(x+4\right)}^2=12\left(4-1\right)$

${\left(x+4\right)}^2=36$

$x+4=\pm 6$

$x=-10; x=2$

Hence $\left(-10, 4\right)$ and $\left(2, 4\right)$ defines the latus rectum.

To determine

To graph: The parabola for the equation ${\left(x+4\right)}^2=12\left(y-1\right)$.

Answer to Problem 36AYU

Explanation of Solution

Given:

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

Calculation:

The graph of the parabola ${\left(x+4\right)}^2=12\left(y-1\right)$.

Using the given information Focus at $\left(-4, 4\right)$; directrix the line $y=-2$, we find Vertex $V\left(-4, 1\right)$ and we have formed the equation and plotted the graph for${\left(x+4\right)}^2=12\left(y-1\right)$.

The focus $F\left(-4, 4\right)$ and the vertex $V\left(-4, 1\right)$ lie on the vertical line $x=-4$ as shown. The equation for directrix D: $y=-2$ is shown as dotted line. The parabola opens up. The points $\left(-10, 4\right)$ and $\left(2, 4\right)$ defines the latus rectum.