Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College identifying a conic section without completing the square. A non degenerate conic section of the form AX squared plus Cy squared plus DX plus EY plus F = 0 in which A and C are not both. 0 is a circle if a = C, a parabola if a * C = 0, and what that really means is a = 0 or b = 0 an ellipse if a doesn't equal C and the multiplication of those two coefficients is positive, and the hyperbola if the multiplication of the A and the C is less than 0 or negative. What happens if both a equals 0 and b = 0? Well, it's no longer conic if a = 0 and b = 0. It's actually just a line. So let's look at some examples. If we have XY squared minus 4X plus two y + 21 = 0, because there's only 1 ^2 term, that tells us that this has to be a parabola. On the next one, we have an X ^2 term that's positive and AY squared term that's negative. When the four times the -9 multiplied together, it gives us a negative value, so we know that this one's going to be a hyperbola. This next one, the coefficients are the same, 4:00 and 4:00, so this one's actually going to be a circle. The next one, nine and four, they're both positive, so when we multiply them together, we get a positive number. The numbers are not the same, so this one is an ellipse. We're going to look at what happens when we rotate our X&Y axis to a new axis called X prime, Y prime through some positive angle Theta. So we're going to put on a triangle, a right triangle on our new rotated axis. So we're going to talk about this triangle here, and because it's a new rotated axis, we're going to think about the distance along the X prime axis as our X distance or our X prime, and the height is going to be our Y prime. So when we think about our trig functions, we can see that we're going to have cosine alpha equaling X prime over R or R Cosine alpha equals X prime. If we thought about sine alpha, that would equal Y prime over R, so our sine alpha would equal Y prime. Now we're going to also look at the original triangle that we could make before the rotation occurred. So this is just our plane X distance and Y distance because it's not on the rotation axis. So we can see that cosine of Theta plus alpha is going to equal X / r. If we cross multiply, we're going to get R cosine Theta cosine alpha minus R sine Theta sine alpha equaling X. Now from above, we showed a moment ago that R cosine alpha is really just X prime. We also show that R sine alpha is really just Y prime. So what we're doing is we're making a relationship X prime cosine Theta minus Y prime. Sine Theta equals the original X. So if I know my angle of rotation and I know my distance is X prime, X prime Y prime, I can find the original X doing the same thing. Now with sine we get sine Theta plus alpha is going to equal y / r So we're going to have R sine Theta cosine alpha plus R cosine Theta sine alpha equaling Y. Now a moment ago we showed that R cosine Theta cosine alpha is really X prime. We also showed that our sine alpha is really Y prime. So when we substitute here we get X prime sine Theta plus Y prime cosine Theta equaling Y. So we have a relationship now X prime cosine Theta minus Y prime, sine Theta equal, X&X prime sine Theta plus Y prime cosine Theta equals Y. Looking at XY equal -4 with the rotation of Theta being 45°, we know that X is going to equal X prime cosine 45 -, y prime sine 45 and that Y is going to equal X prime sine 45 + y prime cosine 45. So we know that cosine 45 and sine 45 is just root 2 / 2. So we get X equal root 2 / 2 X prime minus root 2 / 2 Y prime, and we know that Y equal root 2 / 2 X prime plus root 2 / 2 Y prime. Now, if we substitute those both back into the original equation, we'd get root 2 / 2 X prime minus root 2 / 2 Y prime times root 2 / 2 X prime plus root 2 / 2 Y prime equaling -4. When we foil this out root 2 / 2 times, root 2 / 2 is going to give us 2 fourths or 1/2 X prime. The outer and inner terms are going to cancel and we're going to get -1 half Y prime squared and yx prime squared equaling -4. If we then divided everything through by -4, we would end up with Y prime squared over 8 minus X prime squared over 8, equaling 1. So this is going to be a hyperbola. If we thought about our 45° rotation, it's going to be going in the Y prime direction. So we're going to go up sqrt 8, which will be almost 3. For our vertex. We're going to go down square roots of eight, just almost 3. For our other vertex, we're going to go right sqrt 8 and left sqrt 8. If we put in our box and we connect the corners of the asymptotes, the corners are actually going to go through the original X&Y axis, so we can see that this hyperbola at a 45° rotation is going to look like that. The amount of rotation formula. The general second degree equation of AX squared plus bxy plus Cy squared plus DX plus EY plus F = 0 where B doesn't equal 0 can be rewritten as an equation in X prime and Y prime without an X prime Y prime term by rotating the axis through an angle Theta where cotangent 2 Theta equal a -, C all divided by B. So we're going to do substitution where every X that we see is going to equal X prime cosine Theta minus Y prime sine Theta, and every Y that we see is going to equal X prime sine Theta plus Y prime cosine Theta. So we're going to get A times the X ^2 using that substitution, B times the X and the Y using the substitution. So that's what we get for the first equation that's in blue. To get to the next equation, we're going to actually foil things out and distribute. So we're going to foil out the squared terms or the binomial times binomial. To get the next grouping of the problem, we're going to distribute out the AB and C. Then we're going to group together all the X prime squared coefficients. And then we're going to group together all the X prime Y prime coefficients, all the Y prime squared coefficients, all the X prime coefficients, all the Y prime coefficients, and then finally the F. So we want the X prime Y prime term to equal 0. Thus we want that coefficient to be equal 0 because 0 * X prime Y prime would be 0. So -2 a cosine Theta sine Theta plus B cosine squared Theta minus B sine squared Theta plus 2C cosine Theta, sine Theta equals 0. If we look, we can see that B is going to come out of the cosine squared Theta minus sine squared Theta. And that's just one of our trig identities for cosine 2 Theta. If we look at the remaining two terms, we can see that there's a two cosine Theta sine Theta in each of those terms, leaving us AC minus A. Well, two cosine Theta, sine Theta is really sine 2 Theta by another trig identity. So I'm going to take one of those terms to the other side. This time I'm going to subtract the sine 2 Theta C -, a I'm going to divide each side by B sine 2 Theta. So on the left hand side I end up with cotangent 2 Theta, and on the right hand side I end up with a -, C all divided by B That's our angle of rotation to figure out what the equivalent generic formula would be. If we wanted to go even further, we could come back here and we could state comments such as I know that if I have a prime X prime squared plus B prime X prime Y prime plus C prime Y prime squared plus D prime X prime plus E prime Y prime plus F prime equals zero. That's just the rotation. So now we could actually pair things up and we could say a prime has to equal the coefficient on the X prime squared or the a cosine squared Theta plus B cosine Theta sine Theta plus C sine squared Theta. We could say that our B prime Theta is going to equal 0 which we just showed. So our C prime Theta would equal. Actually in case it doesn't equal zero, we could have B prime being what we just simplified it to. So B cosine 2 Theta plus C -, a sine 2 Theta, C prime would equal a sine squared Theta minus B cosine Theta sine Theta plus C cosine squared Theta. We could do the same thing with D prime, E prime, and F prime, writing the equation of a rotated conic in standard form. We're going to use the given equation AX squared plus bxy plus Cy squared plus DX plus UY plus F = 0 where B doesn't equal 0 to find the cotangent 2 Theta. Then we're going to use the expression for cotangent 2 Theta to determine Theta or the angle of rotation. Then we're going to substitute Theta into the rotation formulas X equal X prime cosine Theta minus Y prime sine Theta and Y equal X prime sine Theta plus Y prime cosine Theta and simplify. Then we're going to substitute the expressions for X&Y in the given equation or the original equation and simplify. The resulting equation should have no X prime, Y prime term. So we're going to write the equation involving X prime and Y prime in standard form. So an example, if I have X ^2 -, sqrt 3 XY +2 Y squared equal 1, I'm going to know that cotangent two Theta is going to equal 1 - 2 over negative root 3 my a - C / b. So cotangent 2 Theta is going to equal 1 over root 3. So if we know that cotangent is one over root 3, cotangent of two Theta we could think of as root 3 / 3. If it's root 3 / 3, we know two Theta is going to be the angle π thirds. If two Theta is π thirds, then Theta has to be Pi 6. So now we're going to figure out that X = X prime, cosine Pi 6 -, y prime, sine of Pi 6 or cosine of Pi 6 is root 3 / 2, sine of Pi 6 is 1/2. Doing the same thing for our Y, we're going to get Y equaling X prime, sine of Pi 6 + y prime cosine of Pi 6, and that's going to equal 1/2 X prime plus root 3 / 2 Y prime. So now we're going to go to that original equation, and instead of X ^2, we're going to stick in sqrt 3 / 2 X prime -1 half Y prime and square that minus root 3 sqrt 3 / 2 X prime -1 half Y prime times 1/2 X prime plus root 3 / 2 Y prime +2. Instead of the Y ^2, we're going to put in 1/2 X prime plus root 3 / 2 Y prime quantity squared equaling 1. If we multiply this out, we're going to get 3/4 X prime squared minus root 3 / 2 X prime Y prime plus 1/4 Y prime squared -3 fourths X prime squared. -3 Square roots of 3 / 4 X prime Y prime plus sqrt 3 / 4 X prime Y prime plus 3/4 Y prime squared plus 1/2 X prime squared plus root 3X prime Y prime plus three halves Y prime squared equaling 1. Now if we combine our like terms, we're going to end up with 1/2 X prime squared +5 halves Y prime squared equaling one. We could think of this as X prime squared over 2 + y prime squared over 2/5 equaling 1. So we know that this is going to be an equation of an ellipse with an angle of rotation of Pi 6. Two is going to be bigger than our 2/5. So when we rotate this our thetas PI6 right about there we're going to go sqrt 2 to the right, sqrt 2 to the left on the rotation. Putting in my perpendicular that didn't look very perpendicular. Putting in my perpendicular, we can see we're going to go up and down sqrt 2 fifths. So a rough sketch would look something like that, an ellipse that's rotated through the angle Theta equal Pi 6. If cotangent 2 Theta is not a familiar angle, then you're going to use a sketch of cotangent 2 Theta and the trig identities that say sine Theta equal sqrt 1 minus cosine 2, Theta over two, and cosine Theta equals square root 1 plus cosine 2 Theta over two. We're only going to deal with positive angles, so we're not going to deal with the plus and minus of these two trig identities, just the positive portion. When we identify a conic section without a rotation of axes, IE we don't do all the work to come up with the equivalent equation. A non degenerate conic section of the form. AX squared plus bxy plus Cy squared plus DX plus EY plus F = 0 is a parabola if b ^2 - 4 AC equals 0. It's an ellipse or a circle if b ^2 - 4 AC is less than 0, and it's a hyperbola if b ^2 - 4 AC is greater than 0. So looking at a few examples, if I have X ^2 - 3 XY plus y ^2 -, X equaling zero, we're going to use b ^2 - 4 AC, so our b ^2 -3 ^2 - 4 * 1 * 1, so 9 - 4, which is five. That's greater than 0, and hence this is a hyperbola. Now we could actually find that hyperbola in the angle of rotation doing what I did in the previous example. This next one, b ^2 so -18 ^2 - 4 * a * C is going to equal 0. When the b ^2 - 4 AC equals zero, it's a parabola. The next one we're going to have our b ^2 negative sqrt 15 ^2 - 4 * a * C so we're going to have 15 - 16 negative one which is less than 0 and hence this one is an ellipse. Thank you and have a wonderful.