Question

Use Cartesian coordinates to express the equations for the parabolas determined by the conditions specified in Exercises 22–31.

$\text{directrix}y=-6,\text{focus}(2,-8)$

Short answer

The equation is$y=-\frac{(x-2{)}^2}{4}-7$.

Step-by-step solution

Step 1. Given information.

We are given,

$\text{directrix}y=-6\text{, focus}(2,-8)$

Step 2. Distance formula.

Now,

$$\begin{gathered} \text{Let}(x,y)\text{be any point on the parabola.} \\ \text{Now find the distance between the point}(x,y)\text{and the focus}(2,-8)\text{.} \\ \text{Distance}=\sqrt{(2-x{)}^2+(-8-y{)}^2}\left[\text{since}{x}_1=x_0,y_1=y_0,x_2=2,y_2=-8\right] \\ \text{Now the distance between the point and the directrix is}|y+6|\text{.} \\ Therefore, \\ \sqrt{(2-x{)}^2+(-8-y{)}^2}=|y+6| \end{gathered}$$

Step 3. Final answer.

On simplifying the equation,

$$\begin{gathered} {\left(\sqrt{(2-x{)}^3+(-8-y{)}^3}\right)}^2=\left(\right|y+6\left|{)}^2 \\ \right(2-x{)}^2+\left(64+y^2+16y\right)=\left(y^2+12y+36\right) \\ (2-x{)}^2+64+16y=12y+36 \\ (2-x{)}^2+64+16y-64-16y=12y+36-64-16y \\ -4y-28=(2-x{)}^2 \\ -4y-28+28=(2-x{)}^2+28 \\ \frac{-4y}{-4}=\frac{(2-x{)}^2+28}{-4} \\ y=-\frac{(x-2{)}^2}{4}-7 \end{gathered}$$