Question

Find the equation of the parabola described. Find the two points that define the latus rectum, and graph the equation by hand.

Focus is at$\left(0,-3\right)$and vertex at$\left(0,0\right)$.

Short answer

The equation of a parabola is $x^2=-12y$. The two points are $\left(-6,-3\right)$and $\left(6,-3\right)$.

The graph is :

Step-by-step solution

Step 1. Given Information.

The given values be:

Focus is at $\left(0,-3\right)$and vertex is at $\left(0,0\right)$.

Step 2.  Equation of a parabola.

The distance from the vertex $\left(0,0\right)$to the focus $\left(0,-3\right)$is $a=-3$.

The equation of a parabola be

role="math" localid="1646651849327" $$\begin{gathered} x^2=4ay \\ {x}^2=-12y \end{gathered}$$.

Step 3. Latus rectum

The two points that determine the latus rectum by letting $y=-3$.

Then,

$$\begin{gathered} x^2=-12y \\ {x}^2=-12\left(-3\right) \\ {x}^2=36 \\ x=\pm 6 \end{gathered}$$

The points role="math" localid="1646652034258" $\left(6,-3\right)$and$\left(-6,-3\right)$that determines the latus rectum.

Step 4. Graph of an equation.

The graph of a parabola equation $x^2=-12y$.