Q9. - Algebra 2

Question

Write an equation for the parabola having focus $\left(3,8\right)$ and directrix $y=4$. Then draw the graph.

Step-by-Step Solution

Answer

The required equation of the parabola is $y=\frac{1}{8}{\left(x-3\right)}^2+6$.

Step 1. Write down the given information.

The given parabola has focus $\left(3,8\right)$ and directrix $y=4$.

Step 2. Concept used.

If $P\left(x,y\right)$ be any point on parabola having focus $\left(f_1,f_2\right)$ and directrix $y=a$ then:

Distance of point $P\left(x,y\right)$ from focus $\left(f_1,f_2\right)=$ Distance of point $P\left(x,y\right)$ from$\left(x,a\right)$

Step 3. Calculation.

Since the given parabola has focus $\left(3,8\right)$ and directrix $y=4$. Therefore, apply the concept stated above,

Distance of point $P\left(x,y\right)$ from focus $\left(3,8\right)=$ Distance of point $P\left(x,y\right)$ from $\left(x,4\right)$

 $\begin{array}{c}\sqrt{{\left(x-3\right)}^2+{\left(y-8\right)}^2}=\sqrt{{\left(x-x\right)}^2+{\left(y-4\right)}^2}\\ {\left(x-3\right)}^2+{\left(y-8\right)}^2={\left(x-x\right)}^2+{\left(y-4\right)}^2\mathrm{....}\left(\text{Squaring}\right)\\ {\left(x-3\right)}^2={\left(y-4\right)}^2-{\left(y-8\right)}^2\\ {\left(x-3\right)}^2=\left(2y-12\right)\left(4\right)\\ {\left(x-3\right)}^2=8y-48\\ y=\frac{1}{8}{\left(x-3\right)}^2+6\end{array}$

Hence, $y=\frac{1}{8}{\left(x-3\right)}^2+6$ is the required equation of the parabola.

Step 4. Sketch the graph of the parabola.

The graph of the parabola $y=\frac{1}{8}{\left(x-3\right)}^2+6$ is shown below.

Step 5. Conclusion.

The required equation of the parabola is $y=\frac{1}{8}{\left(x-3\right)}^2+6$.