Chapter 2.3, Problem 8E

To determine

The equation of parabola using distance formula.

Answer to Problem 8E

The required equation of parabola is $x^2=20y$ .

Explanation of Solution

Given information : Vertex of the parabola is $V\left(0,0\right)$ and focus is $\left(0,5\right)$ .

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.

Since, vertex of the parabola is $V\left(0,0\right)$ and focus is $\left(0,5\right)$ .

Therefore, directrix of parabola is $y=-5$ and arbitrary point on the directrix is $D\left(x,-5\right)$ .

Let, arbitrary point the parabola be $P\left(x,y\right)$ .

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

  $P$ and $D$ .

  $\begin{array}{c}FP=PD\\ \sqrt{{\left(x-0\right)}^2+{\left(y-5\right)}^2}=\sqrt{{\left(x-x\right)}^2+{\left(-5-y\right)}^2}\\ \sqrt{x^2+{\left(y-5\right)}^2}=\sqrt{0+{\left(y+5\right)}^2}\\ x^2+{\left(y-5\right)}^2={\left(y+5\right)}^2\text{ Squaring both side}\\ x^2+y^2-10y+25=y^2+10y+50\\ x^2=10y+10y\\ x^2=20y\end{array}$

Thus, required equation of parabola is $x^2=20y$ .