Chapter 9, Problem 4CT

To determine

To find: Equation of parabola.

Answer to Problem 4CT

  ${\left(y+2\right)}^2=8\left(x-6\right)$

Explanation of Solution

Given information:

Focus of parabola = $\left(8,-2\right)$

Directrix of parabola is $x=4$ .

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

Find the distance between $\left(x,y\right)$ and Focus.

Then Find the distance between $\left(x,y\right)$ and directrix.

Equate these two distances and we will get the equation of parabola in term of $x,y$ .

Distance between $\left(x,y\right)$ and focus is

  $\sqrt{{\left(x-8\right)}^2+{\left(y-\left(-2\right)\right)}^2}=\sqrt{{\left(x-8\right)}^2+{\left(y+2\right)}^2}$

Distance between $\left(x,y\right)$ and Directrix is $\left|x-4\right|$ .

On equating distances

We get

  $\sqrt{{\left(x-8\right)}^2+{\left(y+2\right)}^2}=\left|x-4\right|$

Squaring on both sides

We get

  $\begin{array}{l}\Rightarrow {\left(\sqrt{{\left(x-8\right)}^2+{\left(y+2\right)}^2}\right)}^2={\left(\left|x-4\right|\right)}^2\\ \Rightarrow {\left(x-8\right)}^2+{\left(y+2\right)}^2=x^2-8x+16\\ \Rightarrow x^2-16x+64+y^2+4y+4=x^2-8x+16\\ \Rightarrow x^2-16x+64+y^2+4y+4-x^2+8x-16=0\\ \Rightarrow y^2+4y+4-8x+48=0\\ \Rightarrow {\left(y+2\right)}^2=8x-48\\ \Rightarrow {\left(y+2\right)}^2=8\left(x-6\right)\end{array}$

Therefore

Standard form of Parabola is

  ${\left(y+2\right)}^2=8\left(x-6\right)$

Graph of parabola is