13.1: Analytic Geometry (or...) Conic Sections
Name: _________________________
Distance
- Draw the points \begin{align*}(2,5)\end{align*}
(2,5) , \begin{align*}(10,5)\end{align*}, and \begin{align*}(10,1)\end{align*}, and the triangle they all form. - Find the distance from \begin{align*}(2,5)\end{align*} to \begin{align*}(10,5)\end{align*} (just by looking at it).
- Find the distance from \begin{align*}(10,5)\end{align*} to \begin{align*}(10,1)\end{align*} (just by looking at it).
- Find the distance from \begin{align*}(2,5)\end{align*} to \begin{align*}(10,1)\end{align*}, using your answers to \begin{align*}(2)\end{align*} and \begin{align*}(3)\end{align*} and the Pythagorean Theorem.
- I start at Raleigh Charter High School. I drive \begin{align*}5\;\mathrm{miles}\end{align*} West, and then \begin{align*}12\;\mathrm{miles}\end{align*} North. How far am I now from RCHS? (Hint: the answer is not \begin{align*}17\;\mathrm{miles}\end{align*}. Draw it!)
- Draw a point anywhere in the first quadrant. Instead of labeling the specific coordinates of that point, just label it \begin{align*}(x,y)\end{align*}.
- How far down is it from your point to the \begin{align*}x\end{align*}-axis? (Hint: Think about specific points—such as \begin{align*}(4,2)\end{align*} or \begin{align*}(1,10)\end{align*}—until you can see the pattern and answer the question for the general point \begin{align*}(x,y)\end{align*}.)
- How far across is it from your point to the \begin{align*}y\end{align*}-axis?
- Find the distance \begin{align*}d\end{align*} from the origin to that point \begin{align*}(x,y)\end{align*}, using the Pythagorean Theorem. This will give you a formula for the distance from any point, to the origin.
- Find the distance from the point \begin{align*}(3,7)\end{align*} to the line \begin{align*}y=2\end{align*}.
- Find the distance from the generic point \begin{align*}(x,y)\end{align*} (as before) to the line \begin{align*} y=2\end{align*}.
- Find the distance from the point \begin{align*}(3,7)\end{align*} to the line \begin{align*}y=-2\end{align*}.
- Find the distance from the generic point \begin{align*}(x,y)\end{align*} to the line \begin{align*}y=-2\end{align*}.
- I’m thinking of a point which is exactly \begin{align*}5\;\mathrm{units}\end{align*} away from the point \begin{align*}(0,0)\end{align*}. The \begin{align*}y\end{align*}-coordinate of my point is \begin{align*}0\end{align*}. What is the \begin{align*}x\end{align*}-coordinate? Draw this point.
- I’m thinking of two points which are exactly \begin{align*}5\;\mathrm{units}\end{align*} away from \begin{align*}(0,0)\end{align*}. The \begin{align*}x\end{align*}-coordinates of both points is \begin{align*}4\end{align*}. What are the \begin{align*}y\end{align*}-coordinates? Draw these points on the same graph that you did in \begin{align*}^{\#}14\end{align*}.
- I’m thinking of two points which are exactly \begin{align*}5\;\mathrm{units}\end{align*} away from \begin{align*}(0,0)\end{align*}. The \begin{align*}x\end{align*}-coordinates of both points is \begin{align*}-4\end{align*}. What are the \begin{align*}y\end{align*}-coordinates? Draw these points on the same graph that you did in \begin{align*}^{\#}14\end{align*}.
Name: ______________________________
Homework: Distance
- Draw a point anywhere. Instead of labeling the specific coordinates of that point, just label it \begin{align*}(x_1,y_1)\end{align*}.
- Draw another point somewhere else. Label it \begin{align*}(x_2,y_2)\end{align*}. To make life simple, make this point higher and to the right of the first point.
- Draw the line going from \begin{align*}(x_1,y_1)\end{align*} to \begin{align*}(x_2,y_2)\end{align*}. Then fill in the other two sides of the triangle
- How far up is it from the first point to the second? (As always, start by thinking about specific numbers—then see if you can generalize.)
- How far across is it from the first point to the second?
- Find the distance \begin{align*}d\end{align*} from \begin{align*}(x_1,y_1)\end{align*} to \begin{align*}(x_2,y_2)\end{align*}, using the Pythagorean Theorem. This will give you a general formula for the distance between any two points.
- Plug in \begin{align*}x_2=0\end{align*} and \begin{align*}y_2=0\end{align*} into your formula. You should get the same formula you got on the previous assignment, for the distance between any point and the origin. Do you?
- Draw a line from \begin{align*}(0,0)\end{align*} to \begin{align*}(4,10)\end{align*}. Draw the point at the exact middle of that line. (Use a ruler if you have to.) What are the coordinates of that point?
- Draw a line from \begin{align*}(-3,2)\end{align*} to \begin{align*}(5,-4)\end{align*}. What are the coordinates of the midpoint?
- Look back at your diagram of a line going from \begin{align*}(x_1,y_1)\end{align*} to \begin{align*}(x_2,y_2)\end{align*}. What are the coordinates of the midpoint of that line?
- Find the distance from the point \begin{align*}(3,7)\end{align*} to the line \begin{align*}x=2\end{align*}.
- Find the distance from the generic point \begin{align*}(x,y)\end{align*} to the line \begin{align*}x=2\end{align*}.
- Find the distance from the point \begin{align*}(3,7)\end{align*} to the line \begin{align*}x=-2\end{align*}.
- Find the distance from the generic point \begin{align*}(x,y)\end{align*} to the line \begin{align*}x=-2\end{align*}.
- Find the coordinates of all the points that have \begin{align*}y\end{align*}-coordinate \begin{align*}5\end{align*}, and which are exactly \begin{align*}10\;\mathrm{units}\end{align*} away from the origin.
- Draw all the points you can find which are exactly \begin{align*}3\;\mathrm{units}\end{align*} away from the point \begin{align*}(4,5)\end{align*}.
Name: __________________________
All The Points Equidistant from a Given Point
1. Draw as many points as you can which are exactly \begin{align*}5\;\mathrm{units}\end{align*} away from \begin{align*}(0,0)\end{align*}, and fill in the shape. What shape is it?
2. Now, let’s see if we can find the equation for that shape. How do we do that? Well, for any point \begin{align*}(x,y)\end{align*} to be on the shape, it must be exactly five units away from the origin. So we have to take the sentence:
The point \begin{align*}(x,y)\end{align*} is exactly five units away from the origin
and translate it into math. Then we will have an equation that describes every point on our shape, and no other points. (Stop for a second and discuss this point, make sure it makes sense.)
OK, but how do we do that?
a. Above is a drawing of our point \begin{align*}(x,y)\end{align*}, \begin{align*}5\;\mathrm{units}\end{align*} away from the origin. On the drawing, I have made a little triangle as usual. How long is the vertical line on the right side of the triangle? Label it in the picture.
b. How long is the horizontal line at the bottom of the triangle? Label it in the picture.
c. Now, all three sides are labeled. Just write down the Pythagorean Theorem for this triangle, and you have the equation for our shape!
d. Now, let’s see if it worked. A few points that are obviously part of our shape—that is, they are obviously \begin{align*}5\;\mathrm{units}\end{align*} away from the origin—are the points \begin{align*}(5,0)\end{align*} and \begin{align*}(4,-3)\end{align*}. Plug them both into your equation from the last part and see if they work.
e. A few points that are clearly not part of our shape are \begin{align*}(1,4)\end{align*} and \begin{align*}(-2,7)\end{align*}. Plug them both into your equation for the shape to make sure they don’t work!
3. OK, that was all the points that were \begin{align*}5\;\mathrm{units}\end{align*} away from the origin. Now we’re going to find an equation for the shape that represents all points that are exactly \begin{align*}3\;\mathrm{units}\end{align*} away from the point \begin{align*}(4,-1)\end{align*}. Go through all the same steps we went through above—draw the point \begin{align*}(4,-1)\end{align*} and an arbitrary point \begin{align*}(x,y)\end{align*}, draw a little triangle between them, label the distance from \begin{align*}(x,y)\end{align*} to \begin{align*}(4,-1)\end{align*} as being \begin{align*}3\end{align*}, and write out the Pythagorean Theorem. Don’t forget to test a few points!
4. By now you probably get the idea. So—without going through all that work—write down the equation for all the points that are exactly \begin{align*}7\;\mathrm{units}\end{align*} away from the point \begin{align*}(-5,3)\end{align*}.
5. And finally, the generalization as always: write down the equation for all the points that are exactly \begin{align*}r\;\mathrm{units}\end{align*} away from the point \begin{align*}(h,k)\end{align*}.
Name: \begin{align*}\underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\end{align*}
Circles
Homework: Circles
(*aka “All the Points Equidistant from a Given Point”)
1. Write down the equation for a circle with center \begin{align*}(-3,-6)\end{align*} and radius \begin{align*}9\end{align*}. (Although the wording is different, this is exactly like the problems you did on the in-class assignment.)
2. Now, let’s take it the other way. \begin{align*}(x-4)^2 + (y+8)^2=49\end{align*} is the equation for a circle.
a. What is the center of the circle?
b. What is the radius?
c. Draw the circle.
d. Find two points on the circle (by looking at your drawing) and plug them into the equation to make sure they work. (Show your work!)
\begin{align*}2x^2 + 2y^2 + 8x +24y+100=0\end{align*} is also the equation for a circle. But in order to graph it, we need to put it into our canonical form \begin{align*}(x-h)^2+(y-k)^2=r^2\end{align*}. In order to do that, we have to complete the square...twice! Here’s how it looks.
\begin{align*}2x^2+2y^2+8x+24y+60 & =0 && \text{The original problem} \\ x^2+y^2+4x+12y+30 & =0 && \text{Divide by the coefficient of} \ x^2 \ \text{and} \ y^2 \\ (x^2 +4x)+(y^2 +12y) & =-30 && \text{Collect} \ x \ \text{and} \ y \ \text{terms together, and bring the number to the other side} \\ (x^2+4x\underline{+4})+(y^2+12y\underline{+36}) & =-30\underline{+4+36} && \text{Complete the square in} \ \underline{\text{both}} \ \text{parentheses} \\ (x+2)^2+(y+6)^2 & =10 && \text{Done! The center is} \ (-2, -6) \ \text{and the radius is} \ \sqrt{10}\end{align*}
Got it? Now you try!
3. \begin{align*}3x^2+3y^2+18x+30y-6=0\end{align*}
a. Complete the square—as I did above—to put this into the form:
\begin{align*}(x-h)^2 +(y-k)^2=r^2\end{align*}.
b. What are the center and radius of the circle?
c. Draw the circle.
d. Find two points on the circle (by looking at your drawing) and plug them into the original equation to make sure they work. (Show your work!)
Name: ___________________
All the Points Equidistant from a Point and a Line
On the drawing below is the point \begin{align*}(0,3)\end{align*} and the line \begin{align*}(y=-3)\end{align*}. What I want you to do is to find all the points that are the same distance from \begin{align*}(0,3)\end{align*} that they are from the line \begin{align*}(y=-3)\end{align*}.
One of the points is very obvious. You can get two more of them, exactly, with a bit of thought.
After that you have to start playing around. Feel free to use some sort of measuring device (such as your fingernail, or a pencil eraser). When you think you have the whole shape, call me and let me look.
Name: ___________________
Parabolas
Homework: Vertical and Horizontal Parabolas
1. \begin{align*}y=3x^2 - 30x-70\end{align*}
a. Put into the standard form of a parabola.
b. Vertex:
c. Opens (up / down / right / left):
d. Graph it
2. \begin{align*}x=y^2 +y\end{align*}
a. Put into the standard form of a parabola.
b. Vertex:
c. Opens (up / down / right / left):
d. Graph it
3. Find the equation for a parabola that goes through the points \begin{align*}(0,2)\end{align*} and \begin{align*}(0,8)\end{align*}.
Name: \begin{align*}\underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\end{align*}
Parabolas: From Definition to Equation
We have talked about the geometric definition of a parabola: “all the points in a plane that are the same distance from a given point (the focus) from a given line (the directrix).” And we have talked about the general equations for a parabola:
Vertical parabola: \begin{align*}y=a(x-h)^2 +k\end{align*}
Horizontal parabola: \begin{align*}x=a(y-k)^2+h\end{align*}
What we haven’t done is connect these two things—the definition of a parabola, and the equation for a parabola. We’re going to do it the exact same way we did it for a circle—start with the geometric definition and turn it into an equation.
In the drawing above, I show a parabola whose focus is the origin \begin{align*}(0,0)\end{align*} and directrix is the line \begin{align*}y=-4\end{align*}. On the parabola is a point \begin{align*}(x,y)\end{align*} which represents any point on the parabola.
- \begin{align*}d1\end{align*} is the distance from the point \begin{align*}(x,y)\end{align*} to the focus \begin{align*}(0,0)\end{align*}. What is \begin{align*}d1\end{align*}?
- \begin{align*}d2\end{align*} is the distance from the point \begin{align*}(x,y)\end{align*} to the directrix \begin{align*}(y=-4)\end{align*}. What is \begin{align*}d2\end{align*}?
- What defines the parabola as such—what makes \begin{align*}(x,y)\end{align*} part of the parabola—is that these two distances are the same. Write the equation \begin{align*}d1=d2\end{align*} and you have the parabola.
- Simplify your answer to \begin{align*}^{\#}3\end{align*}; that is, rewrite the equation in the standard form.
- What does your equation say the vertex should be? Does it match the drawing?
Name: ________________________
Sample Test: Distance, Circles and Parabolas
1. Below are the points \begin{align*}(-2,4)\end{align*} and \begin{align*}(-5,-3)\end{align*}.
a. How far is it across from one to the other (the horizontal line in the drawing)?
b. How far is it down from one to the other (the vertical line in the drawing)?
c. How far are the two points from each other?
d. What is the midpoint of the diagonal line?
2. What is the distance from the point \begin{align*}(-1024,3)\end{align*} to the line \begin{align*}y=-1\end{align*}?
3. Find all the points that are exactly \begin{align*}4\;\mathrm{units}\end{align*} away from the origin, where the \begin{align*}x\end{align*} and \begin{align*}y\end{align*} coordinates are the same. (If you get stuck, ask for a hint—it will cost you points, but it’s better than nothing...)
4. \begin{align*}2x^2 +2y^2-6x+4y+2=0\end{align*}
a. Put this equation in the standard form for a circle.
b. What is the center?
c. What is the radius?
d. Graph it on the graph paper.
e. Find one point on your graph, and test it in the original equation. (No credit unless I can see your work!)
5. \begin{align*}x=-\frac{1}{4}y^2+y+2\end{align*}
a. Put this equation in the standard form for a parabola.
b. What direction does it open in?
c. What is the vertex?
d. Graph it on the graph paper.
6. Find the equation for a circle where the center is the point \begin{align*}(-2,5)\end{align*} and the radius is \begin{align*}3\end{align*}.
7. We’re going to find the equation of a parabola whose focus is \begin{align*}(3,2)\end{align*} and whose directrix is the line \begin{align*}x=-3\end{align*}. But we’re going to do it straight from the definition of a parabola.
In the drawing above, I show the focus and the directrix, and an arbitrary point \begin{align*}(x,y)\end{align*} on the parabola.
a. \begin{align*}d1\end{align*} is the distance from the point \begin{align*}(x,y)\end{align*} to the focus \begin{align*}(3,2)\end{align*}. What is \begin{align*}d1\end{align*}?
b. \begin{align*}d2\end{align*} is the distance from the point \begin{align*}(x,y)\end{align*} to the directrix \begin{align*}(x=-3)\end{align*}. What is \begin{align*}d2\end{align*}?
c. What defines the parabola as such—what makes \begin{align*}(x,y)\end{align*} part of the parabola—is that \begin{align*}d1=d2\end{align*}. Write the equation for the parabola.
d. Simplify your answer to part (c); that is, rewrite the equation in the standard form.
Name: \begin{align*}\underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\end{align*}
Distance to This Point Plus Distance to That Point is Constant
On the drawing below are the points \begin{align*}(3,0)\end{align*} and \begin{align*}(-3,0)\end{align*}. We’re going to draw yet another shape—not a circle or a parabola or a line, which are the three shapes we know about. In order to be on our shape, the point \begin{align*}(x,y)\end{align*} must have the following property:
The distance from \begin{align*}(x,y)\end{align*} to \begin{align*}(3,0)\end{align*}, plus the distance from \begin{align*}(x,y)\end{align*} to \begin{align*}(-3,0)\end{align*}, must equal \begin{align*}10\end{align*}.
We’re going to start this one the same way we did our other shapes: intuitively. Your object is to find all the points that have that particular property. Four of them are...well, maybe none of them are exactly obvious, but there are four that you can get exactly, with a little thought. After that, you have to sort of figure it out as we did before.
When you think you know the shape, don’t call it out! Call me over and I will tell you if it’s right.
Name: \begin{align*}\underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\end{align*}
Ellipses
Homework: Ellipses
1. In class, we discussed how to draw an ellipse using a piece of cardboard, two thumbtacks, a string, and a pen or marker. Do this. Bring your drawing in as part of your homework. (Yes, this is a real part of your homework!)
2.\begin{align*}\frac{(x- 2)^2}{9} + \frac{y^2}{\frac{1}{4}}=1\end{align*}
a. Is it horizontal or vertical?
b. What is the center?
c. What is \begin{align*}a\end{align*}?
d. What is \begin{align*}b\end{align*}?
e. What is \begin{align*}c\end{align*}?
f. Graph it.
3. \begin{align*}\frac{4x^2}{9}+25y^2=1\end{align*}
This sort of looks like an ellipse in standard form, doesn’t it? It even has a \begin{align*}1\end{align*} on the right. But it isn’t. Because we have no room in our standard form for that \begin{align*}4\end{align*} and that \begin{align*}25\end{align*}—for numbers multiplied by the \begin{align*}x^2\end{align*} and \begin{align*}y^2\end{align*} terms. How can we get rid of them, to get into standard form, while retaining the \begin{align*}1\end{align*} on the right?
a. Rewrite the left-hand term, \begin{align*}\frac{4x^2}{9}\end{align*}, by dividing the top and bottom of the fraction by \begin{align*}4\end{align*}. Leave the bottom as a fraction, don’t make it a decimal.
b. Rewrite the right-hand term, \begin{align*}25y^2\end{align*}, by dividing the top and bottom of the fraction by \begin{align*}25\end{align*}. Leave the bottom as a fraction don’t make it a decimal.
c. Now, you’re in standard form. What is the center?
d. How long is the major axis?
e. How long is the minor axis?
f. What are the coordinates of the two foci?
g. Graph it.
4. \begin{align*}18x^2 + \frac{1}{2}y^2+108x+5y+170=0\end{align*}
a. Put in standard form.
b. Is it horizontal or vertical?
c. What is the center?
d. How long is the major axis?
e. How long is the minor axis?
f. What are the coordinates of the two foci?
g. Graph it.
5. The major axis of an ellipse runs from \begin{align*}(5,-6)\end{align*} to \begin{align*}(5,12)\end{align*}. One focus is at \begin{align*}(5,-2)\end{align*}. Find the equation for the ellipse.
6. The foci of an ellipse are at \begin{align*}(-2,3)\end{align*} and \begin{align*}(2,3)\end{align*} and the ellipse contains the origin. Find the equation for the ellipse.
7. We traditionally say that the Earth is \begin{align*}93\;\mathrm{million \ miles}\end{align*} away from the sun. However, if it were always \begin{align*}93\;\mathrm{million \ miles}\end{align*} away, that would be a circle (right?). In reality, the Earth travels in an ellipse, with the sun at one focus. According to one Web site I found,
There is a \begin{align*}6\%\end{align*} difference in distance between the time when we're closest to the sun (perihelion) and the time when we're farthest from the sun (aphelion). Perihelion occurs on January 3 and at that point, the earth is \begin{align*}91.4\;\mathrm{million \ miles}\end{align*} away from the sun. At aphelion, July 4, the earth is \begin{align*}94.5\;\mathrm{million \ miles}\end{align*} from the sun.
(\begin{align*}^*\end{align*}Source: http://geography.about.com/library/weekly/aa121498.htm)
Write an equation to describe the orbit of the Earth around the sun. Assume that it is centered on the origin and that the major axis is horizontal. (\begin{align*}^*\end{align*}Why not? There are no axes in space, so you can put them wherever it is most convenient.) Also, work in units of millions of miles—so the numbers you are given are simply \begin{align*}91.4\end{align*} and \begin{align*}94.5\end{align*}.
The Ellipse: From Definition to Equation
Here is the geometric definition of an ellipse. There are two points called the “foci”: in this case, \begin{align*}(-3,0)\end{align*} and \begin{align*}(3,0)\end{align*}. A point is on the ellipse if the sum of its distances to both foci is a certain constant: in this case, I’ll use \begin{align*}10\end{align*}. Note that the foci define the ellipse, but are not part of it.
The point \begin{align*}(x,y)\end{align*} represents any point on the ellipse. \begin{align*}d1\end{align*} is its distance from the first focus, and \begin{align*}d2\end{align*} to the second.
1. Calculate the distance \begin{align*}d1\end{align*} (by drawing a right triangle, as always).
2. Calculate the distance \begin{align*}d2\end{align*} (by drawing a right triangle, as always).
3. Now, to create the equation for the ellipse, write an equation asserting that the sum of \begin{align*}d1\end{align*} and \begin{align*}d2\end{align*} equals \begin{align*}10\end{align*}.
Now simplify it. We did problems like this earlier in the year (radical equations, the “harder” variety that have two radicals). The way you do it is by isolating the square root, and then squaring both sides. In this case, there are two square roots, so you will need to go through that process twice.
4. Rewrite your equation in \begin{align*}^{\#}3\end{align*}, isolating one of the square roots.
5. Square both sides.
6. Multiply out, cancel, combine, simplify. This is the big step! In the end, isolate the only remaining square root.
7. Square both sides again.
8. Multiply out, cancel, combine, and get it to look like the standard form for an ellipse.
9. Now, according to the “machinery” of ellipses, what should that equation look like? Horizontal or vertical? Where should he center be? What are \begin{align*}a, b\end{align*}, and \begin{align*}c?\end{align*} Does all that match the picture we started with?
Name: _________________________
Distance to This Point Minus Distance to That Point is Constant
On the drawing below are the points \begin{align*}(5,0)\end{align*} and \begin{align*}(-5,0)\end{align*}. We’re going to draw yet another shape—our final conic section. In order to be on our shape, the point \begin{align*}(x,y)\end{align*} must have the following property:
Take the distance from \begin{align*}(x,y)\end{align*} to \begin{align*}(5,0)\end{align*}, and the distance from \begin{align*}(x,y)\end{align*} to \begin{align*}(-5,0)\end{align*}. Those two distances must differ by \begin{align*}6\end{align*}. (In other words, this distance minus that distance must equal \begin{align*}\pm 6\end{align*}.)
We’re going to start this one the same way we did our other shapes: intuitively. Your object is to find all the points that have that particular property. Start by finding the two points on the \begin{align*}x-\end{align*}axis that work. After that, you have to sort of work it out as we did before.
When you think you know the shape, don’t call it out! Call me over and I will tell you if it’s right.
Name: ____________________________
Hyperbolas
Homework: Hyperbolas
1. Complete the following chart, showing the similarities and differences between ellipses and hyperbolas.
Ellipse | Hyperbola | |
---|---|---|
How to identify an equation with this shape | Has an \begin{align*}x^2\end{align*} and a \begin{align*}y^2\end{align*} with different coefficients, but the same sign. \begin{align*}3x^2 +2y^2\end{align*} for instance. | |
Equation in standard form: horizontal | \begin{align*}\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2}= 1\end{align*} | |
How can you tell if it is horizontal? | ||
Draw the shape here. Label \begin{align*}a, b\end{align*}, and \begin{align*}c\end{align*} on the drawing. | ||
Center | \begin{align*}(h,k)\end{align*} | |
What \begin{align*}a\end{align*} represents on the graph | ||
What \begin{align*}b\end{align*} represents on the graph | ||
What \begin{align*}c\end{align*} represents on the graph | ||
Which is the biggest, \begin{align*}a, b\end{align*}, or \begin{align*}c?\end{align*} | ||
Mathematical relationship between \begin{align*}a, b\end{align*}, and \begin{align*}c\end{align*}. |
2. \begin{align*}\frac{y^2}{\frac{1}{4}} - \frac{(x-2)^2}{9} =1\end{align*}
a. Is it horizontal or vertical?
b. What is the center?
c. What is \begin{align*}a\end{align*}?
d. What is \begin{align*}b\end{align*}?
e. What is \begin{align*}c\end{align*}?
f. Graph it. Make sure the box and asymptotes can be clearly seen in your graph.
3. \begin{align*}2x^2 +8x-4y^2 +4y=6\end{align*}
a. Put in standard form.
b. Is it horizontal or vertical?
c. What is the center?
d. What is \begin{align*}a\end{align*}?
e. What is \begin{align*}b\end{align*}?
f. What is \begin{align*}c\end{align*}?
g. Graph it. Make sure the box and asymptotes can be clearly seen in your graph.
h. What is the equation for one of the asymptotes that you drew?
4. A hyperbola has vertices at the origin and \begin{align*}(10,0)\end{align*}. One focus is at \begin{align*}(12,0)\end{align*}. Find the equation for the hyperbola.
5. A hyperbola has vertices at \begin{align*}(1,2)\end{align*} and \begin{align*}(1,22)\end{align*}, and goes through the origin.
a. Find the equation for the hyperbola.
b. Find the coordinates of the two foci.
Name: \begin{align*}\underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\end{align*}
Sample Test: Conics 2 (Ellipses and Hyperbolas)
\begin{align*}& && \underline{\text{Horizontal}} && \underline{\text{Vertical}} && \underline{\text{Equations}} \\ & \text{Ellipse} && \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 && \frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2} = 1 && a>b \\ & && && && a^2 = b^2 + c^2 \\ & && \underline{\text{Horizontal}} && \underline{\text{Vertical}} && \underline{\text{Equations}} \\ & \text{Hyperbola} && \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 && \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{a^2} = 1 && c^2 = a^2 + b^2\end{align*}
1. Identify each equation as a line, parabola, circle, ellipse, or hyperbola.
a. \begin{align*}y=\frac{x}{3} \quad \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\end{align*}
b. \begin{align*}y=\frac{3}{x} \quad \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\end{align*}
c. \begin{align*}4y^2=7x+7y+7 \quad \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\end{align*}
d. \begin{align*}5(x+3)^2-5(y+3)^2=9 \quad \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\end{align*}
e. \begin{align*}3x^2+3x+6y+3y^2=4x+7 \quad \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\end{align*}
f. \begin{align*}4x^2+5y^2=4 \quad \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\end{align*}
2. For each shape, is it a function or not? (Just answer yes or no.)
a.Vertical line \begin{align*}\underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\end{align*}
b. Horizontal line \begin{align*}\underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\end{align*}
c. Vertical parabola \begin{align*}\underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\end{align*}
d. Horizontal parabola \begin{align*}\underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\end{align*}
e. Circle \begin{align*}\underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\end{align*}
f. Vertical ellipse \begin{align*}\underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\end{align*}
g. Horizontal ellipse \begin{align*}\underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\end{align*}
h. Vertical hyperbola \begin{align*}\underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\end{align*}
i. Horizontal hyperbola \begin{align*}\underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\end{align*}
3. The United States Capitol building contains an elliptical room. It is \begin{align*}96\;\mathrm{feet}\end{align*} in length and \begin{align*}46\;\mathrm{feet}\end{align*} in width.
a. Write an equation to describe the shape of the room. Assume that it is centered on the origin and that the major axis is horizontal.
b. John Quincy Adams discovered that if he stood at a certain spot in this elliptical chamber, he could overhear conversations being whispered at the opposing party leader’s desk. This is because both the desk, and the secret listening spot, where foci of the ellipse. How far was Adams standing from the desk?
c. How far was Adams standing from the edge of the room closest to him?
4. A comet zooms in from outer space, whips around the sun, and zooms back out. Its path is one branch of a hyperbola,with the sun at one of the foci. Just at the vertex, the comet is \begin{align*}10 \;\mathrm{million \ miles}\end{align*} from the center of the hyperbola, and \begin{align*}15 \;\mathrm{million \ miles}\end{align*} from the sun. Assume the hyperbola is horizontal, and the center of the hyperbola is at \begin{align*}(0,0)\end{align*}.
a. Find the equation of the hyperbola.
b. When the comet is very far away from the sun, its path is more or less a line. As you might guess, that line is represented by the asymptotes of the hyperbola. (One asymptote as it comes in, another as it goes out.) Write the equation for the line that describes the path of the comet after it has left the sun and gotten far out of our solar system.
5. \begin{align*}4x^2-36y^2+144y=153\end{align*}
a. Put in standard form.
b. Is it horizontal or vertical?
c. What is the center?
d. How long is the transverse axis?
e. How long is the conjugate axis?
f. What are the coordinates of the two foci?
g. Graph it. I will be looking for the vertices (the endpoints of the transverse axis), and for the asymptotes to be drawn correctly.
Extra credit
Consider a hyperbola with foci at \begin{align*}(-5,0)\end{align*} and \begin{align*}(5,0)\end{align*}. In order to be on the hyperbola, a point must have the following property: its distance to one focus, minus its distance to the other focus, must be \begin{align*}6\end{align*}.
Write the equation for this hyperbola by using the geometric definition of a hyperbola (\begin{align*}3\end{align*} points). Then simplify it to standard form (\begin{align*}2\end{align*} points).
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Feb 23, 2012Last Modified:
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