Circles and other Conic Sections
We have now switched our attention to circles! The equation for a circle centered at the origin with radius r is given by
Let's say that we wanted to mark off a distance t along the unit circle. We can start at the point (1,0) and move counterclockwise if t is positive, and clockwise if t is negative. The position P(x,y) where you end up is known as the TERMINAL POINT DETERMINED BY THE REAL NUMBER t.
We know that for any real number t, there is a point P on the unit circle, such
that the length of the arc AP - where A is the point (1,0) and P is the
terminating point - is t. Also, if P is any point on the unit circle then the
length of arc AP measured in a counterclockwise direction starting a A is the
unique nonnegative number t. Since the circumference of the unit circle is
2, we see that
We note that the unit circle it symmetric with respect to the x-axis, the
y-axis, the line
|t||Terminal Point determined by t|
Here's is a diagram of the unit circle with these terminal points plotted.
Now from the symmetries of the unit circle, we see that to find a terminal
point in any quadrant, we only need to know the "corresponding" terminal point
in the first quadrant. The reference number T is the shortest distance along
the unit circle between the terminal point determined by t and the x - axis.
To find the reference number it helps if you know the quadrant in which the
terminal point determined by t lies. Once you have that, you can carry on with
the following steps:
- Step 1:
- Find the reference number T.
- Step 2:
- Find the terminal point Q(x,y) determined by t.
- Step 3:
- The terminal point determined by t is P(x, y), where the signs are chosen according to the quadrant in which this terminal point lies.
Although the unit circle is very important, not all circles need to be centered at the origin, nor does the radius need to be 1. If the center of the circle is at the point (h, k), and the radius is any distance, r, then the equation of the circle is: (x - h)2 + (y - k)2 = r2.
Here are some examples.
An ellipse is a squashed circle. To acheive this squashing effect, you can scale the y variable, in a regular equation for a circle, by 2 (or any number). That is, we replace y by 2y to get x2 + (2y)2 = 16, or x2 + 4y2 = 16 (where the equation of the circle was x2 + y2 = 16). You can also scale the x variable in the same way to obtain an ellipse. If we divide both sides of the equation x2 + 4y2 = 16 by 16 then we get:
x2 y2 x2 y2 ---- + ---- = 1 or ---- + ---- = 1 16 4 42 22Thus, the standard form for an equation of an ellipse is generally:
(x - h)2 (y - k)2 ------- + ------- = 1 (given a 0 and b 0 ) a2 b2with the center at (h, k). Here is a picture of an ellipse. An ellipse of this form has two (2) axes; one is the straight line from (-a, 0) to
Here are some examples.
What happens to the curve represented by Ax2 + By2 + Cx + Dy + E = 0 when either A or B is 0?? If B = 0 , but
A C E y = - ---x2 - ---x - --- D D DThis is a quadratic equation - it should look familiar (Polynomial and Root section) - whose graph is a parabola. It looks like half of an ellipse that continues on forever. It opens either upward or downward, depending on the sign of the coefficient of x2, when graphing it.
Similarily, if A = 0, then solving for x would yeild:
B D E x = - ---y2 - ---y - --- C C CThis is also a quadratic but with the usual roles of x and y switched. It is still a parabola but in this case it either opens to the right or the left depending on the sign of the coefficient of y2.
Renaming the constants, we get two standared forms for equations whose solution curves are parabolas:
y = Ax2 + Bx + C Parabola opens up if A > 0 Parabola opens down if A < 0 x = Ay2 + By + C Parabola opens right if A > 0 Parabola opens left if A < 0When sketching the graph of the parabola, the most important point to locate is its vertex which is the "turn-around" point.
If y = Ax2 + Bx + C with A 0, then its vertex is located at (-B/(2A), C-B2/(4A)).
If x = Ay2 + By + C , with A 0, then its vertex is located at (C-B2/(4A), -B/(2A)).
Here are some examples.
Recall what the standard form of an ellipse, centered at the origin, looks like. A hyperbola has the same form except that one of the terms on the left hand side is negative. For example,
x2 y2 - ---- + ---- = 1 a2 b2Solving for y results in:
There is perfect symmetry about the x-axis. The curve has a branch which lies above the x-axis. The other branch is simply the reflection through the x-axis of the top branch. Also note that is smallest when
By reflection, the lower branch has its maximum point at (0, -b) and is asymptotic to -(b/a)|x| while always staying below it. Here is a picture of
x2 y2 - ---- + ---- = 1 a2 b2with the asymptotes being the dotted lines:
If the coefficient of the y2 term was negative instead of the x2 term, the equation would be:
x2 y2 ---- - ---- = 1 a2 b2The graph would look identicle but the roles of x and y interchanged. The curves are still asymptotic to the straight lines y = (b/a)x and y = -(b/a)x, for large values of |x|.
If it is not centered at the origin, but rather at the point (h, k), then the standard equations for a hyperbola are:
(x - h)2 (y - k)2 (x - h)2 (y - k)2 -------- - -------- = 1 and - -------- + -------- = 1 a2 b2 a2 b2Here is an example.
We have seen various curves represented by equations of the form
Here is what can happen:
1) If A = B, then the solution curve is a circle, which may degenerate to a single point or disappear altogether.
2) If A and B have the same sign, then the solution curve is an ellipse, which again may degenerate to a point or disappear.
3) If A and B have opposite signs, then the solution curve is a hyperbola which again may degenerate to two intersecting straight lines.
4) If one of A or B is 0, then the solution curve is a parabola.
5) If A = B = 0, then the solution curve is a straight line.