Q. 26 - Calculus

Question

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

$\text{ directrix }x=x_0\text{, focus }\left(x_1,y_1\right),\text{ where }{x}_0\ne x_1$.

Step-by-Step Solution

Answer

The equation is ${\left(y_1-y\right)}^2=-2x\left(x_0-x_1\right)+x_0^2-x_1^2.$

Step 1. Given information.

The given values are:

$\text{ directrix }x=x_0\text{, focus }\left(x_1,y_1\right),\text{ where }{x}_0\ne x_1$

Step 2. Distance formula.

Let any point on the parabola be $(x,y)$

Co-ordinates of the focus is $(x_1,y_1)$

The distance between the point and the focus is, 

$$\begin{gathered} \text{ Distance }=\sqrt{{\left(x_1-x\right)}^2+{\left(y_1-y\right)}^2}\left[\text{ since }{x}_1=x,y_1=y,x_2=x_1,y_2=y_1\right] \\ \text{ Now the distance between the point and the directrix is }\left|x-x_0\right|\text{. } \\ Therefore, \\ \sqrt{{\left(x_1-x\right)}^2+{\left(y_1-y\right)}^2}=\left|x-x_0\right| \end{gathered}$$

Step 3. Final answer.

On simplifying the equation,

$$\begin{gathered} {\left(\sqrt{{\left(x_1-x\right)}^3+{\left(y_1-y\right)}^3}\right)}^2={\left(\left|x-x_0\right|\right)}^2 \\ {\left(x_1-x\right)}^2+{\left(y_1-y\right)}^2={\left(x-x_0\right)}^2 \\ {x}_1^2-2x{x}_1+x^2+{\left(y_1-y\right)}^2=x^2-2x{x}_0+x_0^2 \\ {x}_1^2-2x{x}_1+{\left(y_1-y\right)}^2=-2x{x}_0+x_0^2 \\ {x}_1^2-2x{x}_1+{\left(y_1-y\right)}^2-x_1^2+2x{x}_1=-2x{x}_0+x_0^2-x_1^2+2x{x}_1 \\ {\left(y_1-y\right)}^2=-2x{x}_0+2x{x}_1+x_0^2-x_1^2 \\ {\left(y_1-y\right)}^2=-2x\left(x_0-x_1\right)+x_0^2-x_1^2 \end{gathered}$$