Question 1

Where is the bomb?

Question

Agent Double OH 3.14 has received a clue from S.A.C. about the coordinates of the bomb. The clue on the fortune cookie reads as follows:

“ Find and graph the reciprocal of the negatively sloped asymptote of the conic shape with this equation:

We intercepted part of the actual coordinates. It is (1, y). Good luck!”

DO NOT MOVE ON UNTIL YOU HAVE ANSWERED THE QUESTION OR

YOU NEED HELP!

Things You Should Know

Here are just a few things which will make explaining this easier when we go over how to get the answer.

This question requires you to do work from two units: Conics and Transformations. First we’ll go over the Conic part.

The Conic Part

The Conic part of this question is all in the equation. We can tell that the Conic shape described in this question is a

hyperbola. How can we tell this? Well let’s look at it!

We know that this is a hyperbola because:• It has an x2 term and a y2 term.• It has denominators under the two terms indicating the lengths of

the semi-transverse axis and the semi-conjugate axis.• The operation being performed is subtraction.

These are the things that help you recognize the equation of a hyperbola.

Great. It’s a Hyperbola. Now what?

Now that we know this is a hyperbola, we need to find its asymptotes. The best way to do this and see them at the same time would be to graph it!

Before we look at graphing it, let’s look at the parts of the equation of a hyperbola.

The Equation of a Hyperbola

This equation represents the standard form of the equation of a hyperbola.

The center of the hyperbola can be found at the coordinates (h,k).

The length of the semi-transverse axis is the same as the value of the parameter under the positive term (in this case, it’s “a”). The transverse axis is equal to 2a.

The length of the semi-conjugate axis is the same as the value of the parameter under the negative term (in this case, it’s “b”). The conjugate axis is equal to 2b.

Which ever term is positive, that is the way the hyperbola opens up (in this case, “x” is positive so it opens along the x-axis or horizontally).

The Equation of a Hyperbola

This equation represents the standard form of the equation of a hyperbola.

The center of the hyperbola can be found at the coordinates (h,k).

The length of the semi-transverse axis is the same as the value of the parameter under the positive term (in this case, it’s “a”). The transverse axis is equal to 2a.

The length of the semi-conjugate axis is the same as the value of the parameter under the negative term (in this case, it’s “b”). The conjugate axis is equal to 2b.

Which ever term is positive, that is the way the hyperbola opens up (in this case, “x” is positive so it opens along the x-axis or horizontally).

Now let’s look at our equation.

The center of this hyperbola is at (6, -4). Remember that in the formula it’s (x-h) and (y-k). That’s why it’s -4 and not +4 in the coordinates.

The length of the semi-transverse axis is 6. Remember that it is a2 that is in the denominator, not a. So square root to find a. The length of the transverse axis is 2a or 12.

The length of the semi-conjugate axis is 8. Remember that it is b2 that is in the denominator, not b. So square root to find b. The length of the conjugate axis is 2b or 16.

But wait………………

The Equation of a Hyperbola

Our equation has it equalling -1 whereas the standard form equation has it equalling 1.

Well there’s an easy way to solve this. Just multiply both sides by -1.

After we do that, we get this equation:

This means that our semi-transverse axis is actually 8 units long and the semi-conjugate axis is actually 6 units long. (The transverse axis length is under the positive term and the conjugate axis length is under the negative term.)

The transverse axis is now 16 units long and the conjugate axis is 12 units longs.

Ok, let’s start graphing this.

The first thing you would graph is the center.

Our center is (6, -4).

Graphing (Conics)

Next we draw the transverse axis.

We draw it as a dotted line because it isn’t

actually part of the graph.

The transverse axis extends 8 units left and right of the center.

Graphing (Conics)

Next we draw the conjugate axis.

We draw it as a dotted line because it isn’t

actually part of the graph.

The conjugate axis extends 6 units up and down from the center.

Graphing (Conics)

Next we draw a box that touches the

endpoints of the conjugate axis and

the transverse axis .

We draw it as a dotted line because it isn’t actually part of the graph.

Graphing (Conics)

See the corners?

Graphing (Conics)

The asymptotes pass through these corners

and the center like so.

Asymptotes are also drawn as dotted lines.

Graphing (Conics)

Now to complete the graph, we have to draw

in the hyperbola. First we draw the vertices

which are at the endpoints of the transverse

axis.

Graphing (Conics)

Now going from each vertex, use the

asymptotes as your guides to draw two

parabolas opening opposite each other.

Remember that your graph should not touch the asymptotes. These lines are put in solid.

Graphing (Conics)

And there you have your graph!

Now it asks for the negatively slopedasymptote. Now we have to find theirequations.

Graphing (Conics)

To find the asymptotes, use the point-slope form.

What point do you use for this? Use the center! And for the slope, it’s rise over run or 3/4.

Graphing (Conics)

For the asymptote running from the bottom-left to the top-right: [P(6,-4), m=3/4]

(y-(-4))=(3/4)(x-6)y+4=(3/4)x-(18/4)y+4=0.75x-4.5y=0.75x-8.5

Graphing (Conics)

So the equation for the positively sloped asymptote is y=0.75x-8.5. But we want the negatively sloped one.

For the asymptote running from the bottom-left to the top-right: [P(6,-4), m=-(3/4)]

Graphing (Conics)

So the equation for the negatively sloped asymptote is y=-0.75x+0.5. We found what we’re looking for!

(y-(-4))=-(3/4)(x-6)y+4=-(3/4)x+(18/4)y+4=-0.75x+4.5y=0.75x+0.5

Now that we have the equation of the negatively sloped asymptote, we can continue on with the second part: changing it to find the reciprocal of the equation.

Transforming

The first thing we have to do is find the invariant points. These are the points on the graph where y=1 and -1. These points are invariant because when you take the reciprocal of 1 and -1, you still get 1 and -1.In this graph, when y=1, x= 2/3 and when y=-1, x=2.

Transforming

After that, we find the root(s). This is the x value when y=0. This line is a vertical asymptote. This is because the reciprocal of 0 is undefined. The root of this equation is at (2/3, 0). This means that the line x=2/3 is an asymptote.

Transforming

Now we begin to draw the graph. First pick an invariant point. We’ll start with the one on the left. Look to the left of the point. If the graph is getting bigger, the reciprocal of that would be it getting smaller. If the graph is getting smaller, the reciprocal of that would be getting bigger.

Transforming

It’s quite simple actually. Try this out.

First write the reciprocals of these numbers that are getting bigger:

1, 2, 3, 4, 5, 6

If you notice, the reciprocals or those numbers got smaller:

1, 1/2, 1/3, 1/4, 1/5, 1/6

The opposite happens with numbers that are getting smaller:

8, 7, 6, 5, 4, 3

The reciprocals of these numbers get bigger:

1/8, 1/7, 1/6, 1/5, 1/4, 1/3

Why Does That Happen?

In this case, going to the left of the left invariant point, the graph gets bigger, so the reciprocal would be getting smaller. When you draw your graph, it should get close to the x-axis but never touch it because that would be when y=0. With a numerator of 1, there is no number of fraction you can put as the denominator to make it equal 0 so it is undefined.

Transforming

Now look between the left invariant point and the asymptote. The graph in this area gets smaller so the reciprocal would get bigger. The reciprocal graph will get close to, but not touch the asymptote.

Transforming

Now the right asymptote. Do the same as the left asymptote. Going left from there towards the asymptote, the graph gets bigger so the reciprocal gets smaller. Remember not to touch the asymptote.

Transforming

Look to the right of the invariant point. The graph there gets smaller so the reciprocal will get bigger. Remember not to touch or cross the x-axis.

Transforming

Congratulations! You found the reciprocal graph! Now we have to find the rest of the coordinates. We know that the x value of the coordinate is 1 so let’s find the y value! There are two ways to do this, the first is by looking on the graph. If you look on the graph, you can see that when x=1, y=-4. However, this may depend on how neatly you drew your graph.

Transforming

To get a more definite answer, it might be better to use the formula. If you recall, it was y=-0.75x+0.5. However, now that it is a reciprocal graph, it is now:

Transforming

Transforming

So let’s plug it in shall we?

Hooray! The coordinates of the next clue are at

(1,-4)!

Highlight under this slideshare

presentation to find the link to episode

2.