Chapter 8.1, Problem 72E

Solution

a.

To graph: The coordinate axes in the construction window.

Given information:

Two the name of axis $x\text{ \& }y$ .

Graph:

  

Interpretation:

Draw/display the coordinate axes (Graph >> Define Coordinate System) as shown:

b.

To graph: The line $y=-1$ as the directrix and the point $\left(0,1\right)$ as the focus.

Given information:

Line $y=-1$ .

The point $\left(0,1\right)$ .

Graph:

  

Interpretation:

Plot point $\left(0,1\right)$ which named $F$ and the line $y=-1$ .

c.

To graph: The horizontal lines and concentric circle.

Given information:

Horizontal line.

Concentric circle.

Graph:

  

Interpretation:

The center of the concentric circles is $\left(0,1\right)$ or point $F$ in our case.

The Snap Points feature will be helpful in constructing the concentric circles and the horizontal lines as shown

d.

To graph: The points where these horizontal lines and concentric circles meet.

Given information:

Construct the points where these horizontal lines and concentric circles meet.

Graph:

  

Interpretation:

Construct the points where these horizontal lines and concentric circles meet .

e.

To prove: The points lie on the parabola with directrix $y=-1$ and focus $\left(0,1\right)$ .

Given information:

The point which is needed to be verified if it lies on the parabola with directrix $y=-1$ and focus $\left(0,1\right)$ .

Proof:

A parabola with directrix $y=-1$ and focus at $(0,1)$ has equation $x^2=4y$ . Since $P$ is on the circle $x^2+(y-$$1{)}^2=n^2$ and on the line $y=n-1$ , its $x$ -coordinate of $P$ must be

  $\begin{array}{c}x=\sqrt{n^2-{((n-1)-1)}^2}\\ =\sqrt{n^2-{(n-2)}^2}\end{array}$

Substituting $\left(\sqrt{n^2-{(n-2)}^2},n-1\right)$ into $x^2=4y$ shows that ${\left(\sqrt{n^2-{(n-2)}^2}\right)}^2=4(n-1)$ where $P$ lies on the parabola $x^2=4y$