Question

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

$\text{directrix}y=0,\text{focus}(0,1)$

Short answer

The equation is$y=\frac{1}{2}{x}^2+\frac{1}{2}$.

Step-by-step solution

Step 1. Given information.

We are given,

$\text{directrix}y=0\text{, focus}(0,1)$

Step 2. Distance Formula.

Now,

$$\begin{gathered} \text{Let}(x,y)\text{be any point on the parabola.} \\ \text{Now find the distance between the point}(x,y)\text{and the focus}(0,1)\text{.} \\ \text{Formula for the distance}=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2} \\ \text{Distance}=\sqrt{(0-x{)}^2+(1-y{)}^2}\left[\text{since}{x}_1=x,y_1=y,x_2=0,y_2=1\right] \end{gathered}$$

Now the distance between the point and the directrix is$\left|y-0\right|$.

Therefore,

$\sqrt{(0-x{)}^2+(1-y{)}^2}=|y-0|$

Step 3. Final answer.

On simplifying the equation,

$$\begin{gathered} {\left(\sqrt{(0-x{)}^3+(1-y{)}^3}\right)}^2=\left(\right|y\left|{)}^2 \\ \right(0-x{)}^2+(1-y{)}^2=y^2 \\ {x}^2+1-2y+y^2=y^2 \\ {x}^2+1-2y+2y=0+2y \\ {x}^2+1=2y \\ \frac{2y}{2}=\frac{x^2+1}{2} \\ y=\frac{1}{2}{x}^2+\frac{1}{2} \end{gathered}$$