Question

determine the vertex, focus, directrix, and endpoints of the latus rectum for the parabola $(y - 2)^2=8(x + 3)$. 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: Identify the standard - form and parameters

The standard form of a parabola opening to the right or left is $(y - k)^2=4p(x - h)$. Comparing $(y - 2)^2 = 8(x + 3)$ with $(y - k)^2=4p(x - h)$, we have $h=-3$, $k = 2$, and $4p=8$, so $p = 2$.

Step2: Find the vertex

The vertex of the parabola $(y - k)^2=4p(x - h)$ is $(h,k)$. Substituting $h=-3$ and $k = 2$, the vertex is $(-3,2)$.

Step3: Find the focus

For a parabola of the form $(y - k)^2=4p(x - h)$ opening to the right ($p>0$), the focus is $(h + p,k)$. Substituting $h=-3$, $p = 2$, and $k = 2$, the focus is $(-3+2,2)=(-1,2)$.

Step4: Find the directrix

The directrix of the parabola $(y - k)^2=4p(x - h)$ is the line $x=h - p$. Substituting $h=-3$ and $p = 2$, we get $x=-3-2=-5$.

Step5: Find the endpoints of the latus - rectum

The length of the latus - rectum is $|4p| = 8$. For the parabola $(y - k)^2=4p(x - h)$, the $x$ - coordinate of the endpoints of the latus - rectum is $x=h + p$. Here, $x=-3 + 2=-1$. Substitute $x=-1$ into the equation $(y - 2)^2=8(x + 3)$. Then $(y - 2)^2=8(-1 + 3)=16$. Solving for $y$, we have $y-2=\pm4$. So $y=2 + 4=6$ and $y=2-4=-2$. The endpoints of the latus - rectum are $(-1,-2)$ and $(-1,6)$.

Answer:

The vertex is $(-3,2)$.
The focus is $(-1,2)$.
The directrix is the line $x=-5$.
The endpoints of the latus rectum are $(-1,-2)$ and $(-1,6)$.