Q. 23 - Calculus

Question

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

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

Step-by-Step Solution

Answer

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

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,

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