Question

determine the vertex, focus, directrix, and endpoints of the latus rectum for the parabola (y^{2}-4y + 8x-12 = 0). enter your answers as points ((a,b)). enter the end points of the latus rectum in increasing order of the (y -)coordinate. the vertex is . the focus is . the directrix is the line . the endpoints of the latus rectum are and

Explanation:

Step1: Rewrite the equation in standard form

Complete the square for the $y$ - terms.

$$
LATEXBLOCK0
$$

The standard - form of a parabola opening to the left is $(y - k)^{2}=4p(x - h)$, where $(h,k)$ is the vertex and $p$ is the distance from the vertex to the focus and from the vertex to the directrix.

Step2: Identify the vertex

Comparing $(y - 2)^{2}=-8(x - 2)$ with $(y - k)^{2}=4p(x - h)$, we have $h = 2$ and $k = 2$. So the vertex $(h,k)$ is $(2,2)$.

Step3: Find the value of $p$

Since $4p=-8$, then $p=-2$.

Step4: Determine the focus

The focus of a parabola of the form $(y - k)^{2}=4p(x - h)$ opening to the left is given by the point $(h + p,k)$. Substituting $h = 2$, $k = 2$, and $p=-2$, we get the focus $(2-2,2)=(0,2)$.

Step5: Find the directrix

The directrix of a parabola of the form $(y - k)^{2}=4p(x - h)$ opening to the left is the line $x=h - p$. Substituting $h = 2$ and $p=-2$, we have $x=2-(-2)=4$.

Step6: Find the endpoints of the latus - rectum

The length of the latus - rectum is $|4p| = 8$. The endpoints of the latus - rectum of a parabola $(y - k)^{2}=4p(x - h)$ opening to the left have the same $x$ - coordinate as the focus, which is $x = 0$.
We substitute $x = 0$ into the original equation $(y - 2)^{2}=-8(0 - 2)=16$. Then $y-2=\pm4$.
If $y-2 = 4$, then $y = 6$; if $y - 2=-4$, then $y=-2$.
The endpoints of the latus - rectum are $(0,-2)$ and $(0,6)$.

Answer:

The vertex is $(2,2)$.
The focus is $(0,2)$.
The directrix is the line $x = 4$.
The endpoints of the latus rectum are $(0,-2)$ and $(0,6)$.