Question

In Problems 4–6, find an equation of the conic described; graph the equation.

Parabola: focus$\left(-1,4.5\right)$, vertex$\left(-1,3\right)$

Short answer

The equation of the graph is ${\left(x+1\right)}^2=6\left(y-3\right)$

The graph of the equation is

Step-by-step solution

Step 1. Given Information

Given to determine an equation of the conic described by

Parabola: focus $\left(-1,4.5\right)$, vertex $\left(-1,3\right)$

The equation is to be graphed.

Step 2. Determining the equation

Both the vertex and focus lie on the line $x=-1$which is the line of symmetry and since the focus lies above vertex, the parabola is an upward opening parabola.

The distance between the vertex and the focus is given by

$$\begin{gathered} a=4.5-3 \\ a=1.5 \end{gathered}$$

The equation of an upward opening parabola is given by ${\left(x-h\right)}^2=4a\left(y-k\right)$where $\left(h.k\right)$is the vertex.

Plugging the values:

$$\begin{gathered} {\left(x-h\right)}^2=4a\left(y-k\right) \\ {\left(x-\left(-1\right)\right)}^2=4\left(1.5\right)\left(y-3\right) \\ {\left(x+1\right)}^2=6\left(y-3\right) \end{gathered}$$

Hence the equation of the parabola is${\left(x+1\right)}^2=6\left(y-3\right)$

Step 3. Graph of the equation

Graphing the equation using graphing utility:

Step 4. Conclusion

The equation of the parabola is ${\left(x+1\right)}^2=6\left(y-3\right)$

The graph of the equation is