Chapter F, Problem 1PS

To determine

Draw a graph of the parabola using given data.

Answer to Problem 1PS

The equations of the parabola are

  $x^2=-8y\text{and}{y}^2=-8x$

Explanation of Solution

Given:

The distance between the focus and the directrix is the $4$ .

Calculation:

A parabola is the set of all points in the plane equidistant from a given point (focus) and a given line (directrix).

The vertex, exactly between the focus and directrix.

Consider a parabola with focus $\left(0,c\right)$ and directrix $y=-c$ , where $c$ is any positive number. Vertex of the parabola is $\left(0,0\right)$ .

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

  $\sqrt{{\left(x-0\right)}^2+{\left(y-c\right)}^2}=\left|y+c\right|$

Squaring both sides.

  $\begin{array}{l}{\left(\sqrt{{\left(x-0\right)}^2+{\left(y-c\right)}^2}\right)}^2={\left(\left|y+c\right|\right)}^2\\ \Rightarrow {\left(x-0\right)}^2+{\left(y-c\right)}^2={\left(y+c\right)}^2\\ \Rightarrow x^2+y^2+c^2-2yc=y^2+c^2+2yc\\ \Rightarrow x^2=4cy\end{array}$

Similarly for negative $c$

  $y^2=4cx$

and for positive $c$ .

  $\begin{array}{l}{x}^2=-4cy\\ y^2=-4cx\end{array}$

In the given problem,

The vertex, exactly between the focus and directrix.

The distance between vertex to directrix $=\frac{4}{2}=2$

Now,

Directrixis $y=2\text{and}x=2$ .

Equation of the parabolas are

  $\begin{array}{l}{x}^2=-4\cdot 2y\Rightarrow x^2=-8y\\ y^2=-4\cdot 2x\Rightarrow y^2=-8x\end{array}$

Write the equation in the standard form of $4f\left(y-k\right)={\left(x-h\right)}^2$

  $4\cdot \left(-2\right)\left(y-0\right)={\left(x-0\right)}^2$

Direction $\to$ it opens to the downward, because the value of ‘ $a$ ’ is negative.

Vertex $=\left(h,k\right)=\left(0,0\right)$

Focus $=\left(0,0+f\right)=\left(0,0-2\right)=\left(0,-2\right)$ .

Directrix,

  $y=0-f=0-\left(-2\right)=2$

Draw a table for the $x$ and $y$ values.

$x$	$0$	$-4$	$4$	$-8$	$8$
$y=-\frac{x^2}{8}$	$0$	$-2$	$-2$	$-8$	$-8$

Hence the graph of the equation is given below.

  

Write the equation in the standard form of $4f\left(x-h\right)={\left(y-k\right)}^2$

  $4\cdot \left(-2\right)\left(x-0\right)={\left(y-0\right)}^2$

Direction $\to$ it opens to the left, because the value of ‘ $a$ ’ is negative.

Vertex $=\left(h,k\right)=\left(0,0\right)$

Focus $=\left(0+f,0\right)=\left(0-2,0\right)=\left(-2,0\right)$ .

Directrix,

  $x=0-f=0-\left(-2\right)=2$

Draw a table for the $x$ and $y$ values.

$x$	$0$	$-2$	$-2$	$-8$	$-8$
$x=-\frac{y^2}{8}$	$0$	$4$	$-4$	$8$	$-8$

Hence the graph of the equation is given below.