Chapter 2.3, Problem 10E

To determine

The equation of parabola using distance formula.

Answer to Problem 10E

The required equation of parabola is $x^2=36y$ .

Explanation of Solution

Given information : Vertex of the parabola is $V\left(0,0\right)$ and equation of directrix is $y=-9$ . 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.

The directrix intersect the line passing through vertex $V\left(0,0\right)$ at $\left(0,-9\right)$ and let focus is $F\left(x,y\right)$ .

First we calculate the focus.

Since, $V\left(0,0\right)$ is midpoint of $\left(x,y\right)$ and $\left(0,-9\right)$ .

Therefore,

  $\left(\frac{x+0}{2},\frac{y-9}{2}\right)=\left(0,0\right)$

Equating this relation, we get

  $x=0$ and $y=9$ .

Therefore, focus of the parabola is $\left(x,y\right)=\left(0,9\right)$ .

Therefore, focus $F\left(0,9\right)$ .

Let, arbitrary point on the parabola be $P\left(x,y\right)$ and a point on directrix be $D\left(x,-9\right)$

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

  $\begin{array}{c}FP=PD\\ \sqrt{{\left(x-0\right)}^2+{\left(y-9\right)}^2}=\sqrt{{\left(x-x\right)}^2+{\left(-9-y\right)}^2}\\ \sqrt{x^2+{\left(y-9\right)}^2}=\sqrt{0+{\left(y+9\right)}^2}\\ x^2+{\left(y-9\right)}^2={\left(y+9\right)}^2\text{ Squaring both side}\\ x^2+y^2-18y+81=y^2+18y+81\\ x^2=18y+18y\\ x^2=36y\end{array}$

Thus, required equation of parabola is $x^2=36y$ .