Chapter 2.3, Problem 4E

To determine

The equation of parabola using distance formula.

Answer to Problem 4E

The required equation of parabola is $x^2=-16y$ .

Explanation of Solution

Given information : Focus of the parabola is $F\left(0,-4\right)$ and equation of directrix is $y=4$ . Also, arbitrary point on directrix and parabola are $D\left(x,4\right)$ and $P\left(x,y\right)$ .

Formula used: Distance between two points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is given by $\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$ .

Since, a parabola is set of all points in a plane equidistant from a fixed point called the focus and a fixed line called the directrix.

Therefore, distance between $F$ and $P$ is equal to distance between $P$ and $D$ .

  $\begin{array}{c}FP=PD\\ \sqrt{{\left(x-0\right)}^2+{\left(y+4\right)}^2}=\sqrt{{\left(x-x\right)}^2+{\left(4-y\right)}^2}\\ \sqrt{x^2+{\left(y+4\right)}^2}=\sqrt{0+{\left(y-4\right)}^2}\\ x^2+{\left(y+4\right)}^2={\left(y-4\right)}^2\text{ Squaring both side}\\ x^2+y^2+8y+16=y^2-8y+16\\ x^2=-8y-8y\\ x^2=-16y\end{array}$

Thus, required equation of parabola is $x^2=-16y$ .