Chapter 10.2, Problem 34AYU

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

To determine

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

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

Answer to Problem 34AYU

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

Explanation of Solution

Given:

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

Formula used:

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

Calculation:

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

The Vertex is at $V\left(-1,4\right)$ and $a=3$

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

Hence the equation of parabola is

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

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

By substituting $a=3$

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

The equation of the parabola is

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

Substituting $x=2$ we get,

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

${\left(y-4\right)}^2=12\left(3\right)$

$y-4=\pm 6$

$y=-2, y=10$

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

To determine

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

Answer to Problem 34AYU

Explanation of Solution

Given:

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

Calculation:

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

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

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