Chapter 8.1, Problem 71E

Solution

a.

To graph: Placing a line $l$ (directrix) and a point $F$(focus) not on the line in the construction window

Given information:

Line $l$

Point $F$(focus)

Graph:

  

Interpretation:

Drew a random $l$ line for the directrix and a point $F$ not on the line.

b.

To graph: Construct a point $A$on the directrix, and then the segment $AF$ .

Given information:

Point $A$ .

Segment $AF$ .

Graph:

  

Interpretation:

Place a point on line $l$ , naming it $A$ and connect $A$ and $F$ ,forming the segment $AF$ .

c.

To graph: Construct a point $P$where the perpendicular bisector of $AF$ meets the line perpendicular to $l$through $A$ .

Given information:

Point $P$ .

The perpendicular bisector of $AF$ .

Graph:

  

Interpretation:

Construct the perpendicular bisector of $AF$ by locating its midpoint first then constructing the perpendicular line through it. Then, construct the perpendicular line to line $l$ through $A$ . The point of intersection of the two lines constructed is point $P$ ..

d.

To graph: What curve does $P$trace out as $A$moves.

Given information:

Point $P$ .

Move point $A$ .

Graph:

  

Interpretation:

Select point $P$ and use the Trace Intersection feature (from Display). Move point $A$ along line $l$ back and forth and you will notice that point $P$ traces out a parabola as shown.

e.

To prove: Prove that your answer parabola is correct.

Given information:

Parabola equation trace.

Proof:

With labels as shown, can express the coordinates of $P$ using the point-slope equation of the line $P M$

$ $\begin{array}{c}y-\frac{l+c}{2}=\frac{x-b}{c-l}\\ x-\frac{x+b}{2}y-\frac{l+c}{2}=\frac{{(x-b)}^2}{2(c-l)}\\ $ 2(c-l)\times y-\frac{l+c}{2}={(x-b)}^2$\end{array}$ $

This is the equation of a parabola with vertex at $b,\frac{l+c}{2}$ and focus at $b,\frac{l+c}{2}+p$ where $p=\frac{c-l}{2}$ .