Q37. - Algebra 2

Question

Write an equation for the parabola having vertex $\left(8,6\right)$ and focus $\left(2,6\right)$. Then draw the graph.

Step-by-Step Solution

Answer

The required equation of the parabola is $x=-\frac{1}{24}{\left(y-6\right)}^2+8$.

Step 1. Write down the given information.

The given parabola has vertex $\left(8,6\right)$ and focus $\left(2,6\right)$.

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.

 $\boxed{\begin{array}{ccc}\text{Form of equations}& y=a{\left(x-h\right)}^2+k& x=a{\left(y-k\right)}^2+h\\ \text{Vertex}& \left(h,k\right)& \left(h,k\right)\\ \text{Axis of symmetry}& x=h& y=k\\ \text{Focus}& \left(h,k+\frac{1}{4a}\right)& \left(h+\frac{1}{4a},k\right)\\ \text{Directrix}& y=k-\frac{1}{4a}& x=h-\frac{1}{4a}\\ \text{Direction of opening}& \begin{array}{l}\text{upward if }a>0,\\ \text{downward if }a<0\end{array}& \begin{array}{l}\text{right if }a>0,\\ \text{left if }a<0\end{array}\\ \text{Length of latus rectum}& \left|\frac{1}{a}\right|\text{units}& \left|\frac{1}{a}\right|\text{units}\end{array}}$

Step 3. Calculation.

From the given vertex $\left(8,6\right)$ and focus $\left(2,6\right)$ 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 $\left(h,k\right)=\left(8,6\right)$ in $x=a{\left(y-k\right)}^2+h$.

 $\begin{array}{l}x=a{\left(y-\left(6\right)\right)}^2+\left(8\right)\\ x=a{\left(y-6\right)}^2+\mathrm{8....}\left(1\right)\end{array}$

Plugging for vertex $\left(h,k\right)=\left(8,6\right)$ in focus $\left(h+\frac{1}{4a},k\right)$ are equating.

 $\begin{array}{c}\left(h+\frac{1}{4a},k\right)=\left(2,6\right)\mathrm{....}\left(\text{Given}\right)\\ \left(8+\frac{1}{4a},6\right)=\left(2,6\right)\\ 8+\frac{1}{4a}=2\\ \frac{1}{4a}=-6\\ a=-\frac{1}{24}\end{array}$

Plugging $a=-\frac{1}{24}$ in (1) gives $x=-\frac{1}{24}{\left(y-6\right)}^2+8$which is the required equation of the parabola.

Step 4. Sketch the graph of the parabola.

The graph of the parabola $x=-\frac{1}{24}{\left(y-6\right)}^2+8$ is shown below.

Step 5. Conclusion.

The required equation of the parabola is $x=-\frac{1}{24}{\left(y-6\right)}^2+8$.