Question

Write an equation for the parabola having focus $\left(-4,-2\right)$and directrix $x=-8$. Then draw the graph.

Short answer

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

Step-by-step solution

Step 1. Write down the given information.

The given parabola has focus $\left(-4,-2\right)$ and directrix $x=-8$.

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 $x=a$ 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 $\left(a,y\right)$

Step 3. Calculation.

Since the given parabola has focus $\left(-4,-2\right)$ and directrix $x=-8$. Therefore, apply the concept stated above,

Distance of point $P\left(x,y\right)$ from focus $\left(-4,-2\right)$ Distance of point $P\left(x,y\right)$ from $\left(-8,y\right)$

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

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

Step 4. Sketch the graph of the parabola.

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

Step 5. Conclusion.

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