Question

question
find the endpoints of the latus rectum of the parabola below.
$(y + 3)^2=(x - 2)$
enter the endpoints as ordered pairs $(x,y)$.
provide your answer below:
the endpoints of the latus rectum are and .

Explanation:

Step1: Identify the form of the parabola

The given equation $(y + 3)^2=(x - 2)$ is of the form $(y - k)^2 = 4p(x - h)$, where $(h,k)$ is the vertex of the parabola. Comparing, we have $h = 2,k=-3$ and $4p = 1$, so $p=\frac{1}{4}$.

Step2: Find the x - coordinate of the endpoints of the latus - rectum

The x - coordinate of the endpoints of the latus - rectum for a parabola of the form $(y - k)^2=4p(x - h)$ is $x=h + p$. Substituting $h = 2$ and $p=\frac{1}{4}$, we get $x=2+\frac{1}{4}=\frac{9}{4}$.

Step3: Find the y - coordinates of the endpoints of the latus - rectum

Substitute $x=\frac{9}{4}$ into the equation $(y + 3)^2=(x - 2)$. So $(y + 3)^2=\frac{9}{4}-2=\frac{1}{4}$. Then $y+3=\pm\frac{1}{2}$.
When $y + 3=\frac{1}{2}$, $y=\frac{1}{2}-3=-\frac{5}{2}$.
When $y + 3=-\frac{1}{2}$, $y=-\frac{1}{2}-3=-\frac{7}{2}$.

Answer:

$(\frac{9}{4},-\frac{5}{2})$ and $(\frac{9}{4},-\frac{7}{2})$