Q. 22 - Calculus

Question

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

$\text{ directrix }x=3\text{, focus }(0,1)$

Step-by-Step Solution

Answer

The equation is $(y-1{)}^2=-6x+9.$

Step 1. Given information.

We are given,

$\text{ directrix }x=3\text{, focus }(0,1)$

Step 2. Distance formula.

Now,

$\text{ Let }(x,y)\text{ be any point on the parabola. }$

We know,

$$\begin{gathered} \text{ Formula for the distance }=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2} \\ \text{ Distance }=\sqrt{(0-x{)}^2+(1-y{)}^2}\left[\text{ since }{x}_1=x,y_1=y,x_2=0,y_2=1\right] \end{gathered}$$

Now, the distance between the point and directrix $\left|x-3\right|$

Then,

$\sqrt{(0-x{)}^2+(1-y{)}^2}=|x-3|$

Step 3. Final answer.

On simplifying the equation,

$$\begin{gathered} {\left(\sqrt{(0-x{)}^2+(1-y{)}^2}\right)}^2=\left(\right|x-3\left|{)}^2 \\ \right(0-x{)}^2+(1-y{)}^2=(x-3{)}^2 \\ {x}^2+1+y^2-2y=\left(x^2-6x+9\right) \\ 1+y^2-2y=-6x+9 \\ {y}^2-2y+1=-6x+9 \\ (y-1{)}^2=-6x+9 \end{gathered}$$