Q. 27 - Calculus

Question

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

$\text{ directrix }y=y_0\text{, focus }\left(x_1,y_1\right)\text{, where }{y}_0\ne y_1$

Step-by-Step Solution

Answer

The equation is $y=\frac{1}{2}·\frac{{\left(x-x_1\right)}^2}{\left(y_1-y_0\right)}+\frac{\left(y_0+y_1\right)}{2}$.

Step 1. Given information.

The given values are,

$\text{ directrix }y=y_0\text{, focus }\left(x_1,y_1\right)\text{, where }{y}_0\ne y_1$

Step 2. Distance formula.

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

Let $(x_1,y_1)$ be the focus.

Therefore, by distance formula,

$$\begin{gathered} \text{ Distance }=\sqrt{{\left(x_1-x\right)}^2+{\left(y_1-y\right)}^2}\left[\text{ since }{x}_1=x,y_1=y,x_2=x_1,y_2=y_1\right] \\ \text{ Now the distance between the point and the directrix is }\left|y-y_0\right|\text{. } \\ Therefore, \\ \sqrt{{\left(x_1-x\right)}^2+{\left(y_1-y\right)}^2}=\left|y-y_0\right| \end{gathered}$$

Step 3. Final answer.

On simplifying the equation,

$$\begin{gathered} {\left(\sqrt{{\left(x_1-x\right)}^3+{\left(y_1-y\right)}^3}\right)}^2={\left(\left|y-y_0\right|\right)}^2 \\ {\left(x-x_1\right)}^2+{\left(y-y_1\right)}^2={\left(y-y_0\right)}^2 \\ {\left(x-x_1\right)}^2=\left(y^2+y_0^2-2y{y}_0\right)-\left(y^2+y_1^2-2y{y}_1\right) \\ {\left(x-x_1\right)}^2=\left(y_0^2-y_1^2\right)-2y\left(y_0-y_1\right) \\ \frac{{\left(x-x_1\right)}^2}{\left(y_0-y_1\right)}=\frac{\left(y_0-y_1\right)\left(\left(y_0+y_1\right)-2y\right)}{\left(y_0-y_1\right)} \\ \frac{{\left(x-x_1\right)}^2}{\left(y_0-y_1\right)}=\left(y_0+y_1\right)-2y \\ or \\ -2y=\frac{{\left(x-x_1\right)}^2}{\left(y_0-y_1\right)}-\left(y_0+y_1\right) \\ \frac{-2y}{-2}=\frac{1}{-2}·\frac{{\left(x-x_1\right)}^2}{\left(y_0-y_1\right)}-\frac{1}{-2}·\left(y_0+y_1\right) \\ y=\frac{1}{2}·\frac{{\left(x-x_1\right)}^2}{\left(y_1-y_0\right)}+\frac{\left(y_0+y_1\right)}{2} \end{gathered}$$