Question

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

Vertex is at$\left(4,-2\right)$and focus is at$\left(6,-2\right)$.

Short answer

The equation of a parabola is

${\left(y+2\right)}^2=8\left(x-4\right)$.

The points are$\left(6,2\right);\left(6,-6\right)$.

The graph of an equation is

Step-by-step solution

Step 1. Given Information.

The given vertex is at$\left(4,-2\right)$and focus is at$\left(6,-2\right)$.

Step 2. Equation of a parabola.

The vertex is at $\left(4,-2\right)$and focus $\left(6,-2\right)$both lie on the horizontal line $y=-2$(the axis of symmetry). The distance a from the vertex to the focus is $a=-2$.

The parabola opens to the right. The form of a equation is

${\left(y-k\right)}^2=-4a\left(x-h\right)$.

where $\left(h,k\right)=\left(4,-2\right)$and $a=-2$. Therefore, the equation is

${\left(y+2\right)}^2=8\left(x-4\right)$.

Step 3. Latus rectum.

The two points that determines the latus rectum by letting $x=6$, so that

$$\begin{gathered} {\left(y+2\right)}^2=8\left(x-4\right) \\ {\left(y+2\right)}^2=8\left(2\right) \\ {\left(y+2\right)}^2=16 \\ y+2=\pm 4 \\ y+2=4,y+2=-4 \\ y=2,-6. \end{gathered}$$

The points are$\left(6,2\right)$and$\left(6,-6\right)$.

Step 4. Graphing Utility.

The graph of a parabola is