Question

question
given the parabola below, find the endpoints of the latus rectum.
(y + 3)^2=-2(x - 2)
select the correct answer below:
the endpoints of the latus rectum are (3/2,-1) and (3/2,-5).
the endpoints of the latus rectum are (3/2,-3) and (3/2,-5).
the endpoints of the latus rectum are (3/2,-2) and (3/2,-4).
the endpoints of the latus rectum are (3/2,-4) and (3/2,-6).
the endpoints of the latus rectum are (3/2,0) and (3/2,-6).
the endpoints of the latus rectum are (3/2,2) and (3/2,-8).
the endpoints of the latus rectum are (3/2,3) and (3/2,-9).

Explanation:

Step1: Identify the form of the parabola

The given equation is \((y + 3)^2=-2(x - 2)\), which is of the form \((y - k)^2 = 4p(x - h)\). Here, \(h = 2\), \(k=-3\), and \(4p=-2\), so \(p=-\frac{1}{2}\).

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 \(p=-\frac{1}{2}\), we get \(x=2-\frac{1}{2}=\frac{3}{2}\).

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

Substitute \(x = \frac{3}{2}\) into the equation \((y + 3)^2=-2(x - 2)\). So \((y + 3)^2=-2(\frac{3}{2}-2)=-2\times(-\frac{1}{2}) = 1\). Then \(y+3=\pm1\). When \(y + 3 = 1\), \(y=-2\); when \(y + 3=-1\), \(y=-4\).

Answer:

The endpoints of the latus rectum are \((\frac{3}{2},-2)\) and \((\frac{3}{2},-4)\).