Question

1 determine the vertex, focus, directrix, and endpoints of the latus rectum for the following parabola $y^{2}=-8x$.
2 determine the vertex, focus, directrix, and endpoints of the latus rectum for the following parabola $(x - 1)^{2}=-12(y - 2)$.
3 determine the vertex, focus, directrix, and endpoints of the latus rectum for the following parabola $y^{2}+4y + 8x-12 = 0$.

Explanation:

Step1: Recall the standard - form of a parabola

The standard form of a parabola of the form $y^{2}=4px$ has vertex at $(0,0)$, focus at $(p,0)$, directrix $x = - p$, and endpoints of the latus - rectum at $(p,\pm2p)$.
For the parabola $y^{2}=-8x$, we can rewrite it as $y^{2}=4(-2)x$.

Step2: Find the vertex

For the form $y^{2}=4px$, the vertex is $(0,0)$.

Step3: Find the focus

Since $4p=-8$, then $p=-2$. The focus is $(p,0)=(-2,0)$.

Step4: Find the directrix

The directrix is $x=-p$, so $x = 2$.

Step5: Find the endpoints of the latus - rectum

Substitute $x = p=-2$ into the equation $y^{2}=-8x$. We get $y^{2}=-8\times(-2)=16$, so $y=\pm4$. The endpoints of the latus - rectum are $(-2,4)$ and $(-2, - 4)$.

for the second parabola $(x - 1)^{2}=-12(y - 2)$:

Step1: Recall the standard - form of a parabola

The standard form of a parabola of the form $(x - h)^{2}=4p(y - k)$ has vertex at $(h,k)$, focus at $(h,k + p)$, directrix $y=k - p$, and endpoints of the latus - rectum at $(h\pm2p,k + p)$.
Here $h = 1,k = 2$, and $4p=-12$, so $p=-3$.

Step2: Find the vertex

The vertex is $(h,k)=(1,2)$.

Step3: Find the focus

The focus is $(h,k + p)=(1,2-3)=(1,-1)$.

Step4: Find the directrix

The directrix is $y=k - p=2-(-3)=5$.

Step5: Find the endpoints of the latus - rectum

$h\pm2p=1\pm2\times(-3)=1\pm(-6)$. The endpoints are $(1 + 6,-1)=(7,-1)$ and $(1-6,-1)=(-5,-1)$.

for the third parabola $y^{2}+4y + 8x-12 = 0$:

Step1: Complete the square for the $y$ - terms

$y^{2}+4y=(y + 2)^{2}-4$. So the equation becomes $(y + 2)^{2}-4+8x-12 = 0$, or $(y + 2)^{2}=-8x + 16$, or $(y + 2)^{2}=-8(x - 2)$.

Step2: Recall the standard - form of a parabola

The standard form is $(y - k)^{2}=4p(x - h)$, where $h = 2,k=-2$, and $4p=-8$, so $p=-2$.

Step3: Find the vertex

The vertex is $(h,k)=(2,-2)$.

Step4: Find the focus

The focus is $(h + p,k)=(2-2,-2)=(0,-2)$.

Step5: Find the directrix

The directrix is $x=h - p=2-(-2)=4$.

Step6: Find the endpoints of the latus - rectum

Substitute $x = h + p=0$ into the original equation. When $x = 0$, we have $y^{2}+4y-12 = 0$. Factoring gives $(y + 6)(y - 2)=0$, so $y=-6$ or $y = 2$. The endpoints are $(0,-6)$ and $(0,2)$.

Answer:

Vertex: $(0,0)$
Focus: $(-2,0)$
Directrix: $x = 2$
Endpoints of the latus rectum: $(-2,4),(-2,-4)$