Question

Write an equation for the parabola having vertex $\left(5,-1\right)$and focus $\left(3,-1\right)$. Then draw the graph.

Short answer

The required equation of the parabola is$x=-\frac{1}{8}{\left(y+1\right)}^2+5$.

Step-by-step solution

Step 1. Write down the given information.

The given parabola has vertex$\left(5,-1\right)$ and focus $\left(3,-1\right)$.

Step 2. Concept used.

For two different forms of equations of parabola stated below, use the following key-concept to find vertex, axis of symmetry, focus, directrix, direction of opening of parabola and length of latus rectum.

$\boxed{\begin{array}{ccc}\text{Formofequations}& y=a{\left(x-h\right)}^2+k& x=a{\left(y-k\right)}^2+h\\ \text{Vertex}& \left(h,k\right)& \left(h,k\right)\\ \text{Axisofsymmetry}& x=h& y=k\\ \text{Focus}& \left(h,k+\frac{1}{4a}\right)& \left(h+\frac{1}{4a},k\right)\\ \text{Directrix}& y=k-\frac{1}{4a}& x=h-\frac{1}{4a}\\ \text{Directionofopening}& \begin{array}{l}\text{upwardif}a>0,\\ \text{downwardif}a<0\end{array}& \begin{array}{l}\text{rightif}a>0,\\ \text{leftif}a<0\end{array}\\ \text{Lengthoflatusrectum}& \left|\frac{1}{a}\right|\text{units}& \left|\frac{1}{a}\right|\text{units}\end{array}}$

Step 3. Calculation.

From the givenvertex $\left(5,-1\right)$ and focus $\left(3,-1\right)$ it can be interpreted that the y-coordinate remains unchanged and x-coordinate changes.

As, x-coordinate changes therefore the equation of the parabola in standard form will be like:

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

Plugging for vertex $\left(h,k\right)=\left(5,-1\right)$in $x=a{\left(y-k\right)}^2+h$.

$\begin{array}{l}x=a{\left(y-\left(-1\right)\right)}^2+\left(5\right)\\ x=a{\left(y+1\right)}^2+\mathrm{5....}\left(1\right)\end{array}$

Plugging for vertex $\left(h,k\right)=\left(5,-1\right)$in focus $\left(h+\frac{1}{4a},k\right)$ are equating.

$\begin{array}{c}\left(h+\frac{1}{4a},k\right)=\left(3,-1\right)\mathrm{....}\left(\text{Given}\right)\\ \left(5+\frac{1}{4a},-1\right)=\left(3,-1\right)\\ 5+\frac{1}{4a}=3\\ \frac{1}{4a}=-2\\ a=-\frac{1}{8}\end{array}$

Plugging$a=-\frac{1}{8}$ in (1) gives $x=-\frac{1}{8}{\left(y+1\right)}^2+5$ which is the required equation of the parabola.

Step 4. Sketch the graph of the parabola.

The graph of the parabola $x=-\frac{1}{8}{\left(y+1\right)}^2+5$ is shown below.

Step 5. Conclusion.

The required equation of the parabola is $x=-\frac{1}{8}{\left(y+1\right)}^2+5$.