Rotate the axes to eliminate the xy-term in the equation. Then write the equation in standard form. 7x2 + 26xy + 7y2 - 24 = 0

6 years agoThis should help you answer this question: http://en.wikipedia.org/wiki/Rotation_of_axes

6 years agoI've looked at that, but I'm not sure how to apply it to my question.

6 years agoIf you read through that web page you'll find it actually shows you the answer to your question.

6 years agoLook at the section titled: Elimination of the xy term by the rotation formula

6 years agoBut I have tried using that formula and either I am not doing it correctly or I do not understand it. That is why I need help.

6 years agowhich part do you not understand?

6 years agoI'm not quite sure... when I tried to plug in the A-C, I got 0, which does not help me to get an answer

6 years agoA-C=0 is the correct approach

6 years agothat will lead to:\[\tan(2\theta)=\infty\]

6 years agotherefore:\[2\theta=?\]

6 years ago∞?

6 years agoif you know plot this hyperbola, it's easier. Look... B^2-4AC=480 (in this case), so we have a hyperbola. A=B=7, so we have \[\theta=\pi/4\].

6 years agowhich angle will give you infinity when you take its tan?

6 years agoi don't know

6 years agohave you studied trigonometry?

6 years agono, this is in my precalc course

6 years agohmmmm... well you do need to understad basic trigonometry in order to do this question

6 years agothat is why i came here. i did not know how to do this type of question and came for help.

6 years agothis might help you understand the basics: http://www.clarku.edu/~djoyce/trig/

6 years agothere are a good set of short online videos on this here as well: http://www.khanacademy.org/math/trigonometry

6 years agoPlease, how book you use?

6 years agoI am using the Larson Hostetler Seventh Edition Precalculus book

6 years ago@netlopes1 , how did you get pi/4 ?

6 years ago@dpaInc do you know how to solve this type of problem?

6 years agoPlease, read these pages....

6 years agoyes.. but i was doing it the way @asnaseer , was doing it... eventually, you should come up with pi/4 as the rotation. but i think there is some other way and that's why i asked @netlopes1 how he came up with that answer.... i think he can explain his method better...

6 years agoSimply, i used this link: http://en.wikipedia.org/wiki/Rotation_of_axes and observe that A=B=7 and this is a hyperbola equation, ok? In these cases the angle is 45º or pi/4.

6 years agoOk, this may (or mostly likely will) get messy. However, this method does work. Let me know if it makes sense or not. Start with the formula given on the wiki page (A*cos^2(theta)+Bsin(theta)*cos(theta)+C*sin^2(theta))*x^2 +(A*sin^2(theta)-Bsin(theta)*cos(theta)+C*cos^2(theta))*y^2 +(D*cos(theta)+E*sin(theta))*x +(-D*sin(theta)+E*cos(theta))*y +F = 0 The formula is broken up into multiple lines (this is one really really big equation). The idea here is to plug in theta = pi/4 and evaluate ------------------------------------------------------- Step 1) Plug in theta = pi/4 (7*cos^2(pi/4)+26*sin(pi/4)*cos(pi/4)+7*sin^2(pi/4))*x^2 +(7*sin^2(pi/4)-26*sin(pi/4)*cos(pi/4)+7*cos^2(pi/4))*y^2 +(0*cos(pi/4)+0*sin(pi/4))*x +(-0*sin(pi/4)+0*cos(pi/4))*y +24 = 0 ------------------------------------------------------- Step 2) Evaluate the sine and cosine of pi/4 (in this case, they both evaluate to 0.707106781186547 approximately) (7*0.5+26*0.707106781186547*0.707106781186548+7*0.5)*x^2 +(7*0.5-26*0.707106781186547*0.707106781186548+7*0.5)*y^2 +(0*0.707106781186548+0*0.707106781186547)*x +(-0*0.707106781186547+0*0.707106781186548)*y +24 = 0 ------------------------------------------------------- Step 3) Square each value (7*0.5+26*0.707106781186547*0.707106781186548+7*0.5)*x^2 +(7*0.5-26*0.707106781186547*0.707106781186548+7*0.5)*y^2 +(0*0.707106781186548+0*0.707106781186547)*x +(-0*0.707106781186547+0*0.707106781186548)*y +24 = 0 ------------------------------------------------------- Step 4) Multiply (3.5+13+3.5)*x^2 +(3.5-13+3.5)*y^2 +(0+0)*x +(0+0)*y +24 = 0 ------------------------------------------------------- Step 5) Combine like terms (20)*x^2 +(-6)*y^2 +(0)*x +(0)*y +24 = 0 ------------------------------------------------------- Step 6) Condense into one line (this is now possible since the equation is much shorter/smaller. 20x^2-6y^2+0x+0y+24 = 0 ------------------------------------------------------- Step 7) Erase the 0x and 0y terms 20x^2-6y^2+24 = 0 ====================================================================================================================== Answer: So 7x^2+26xy+7y^2+24=0 rotates pi/4 radians (or 45 degrees) clockwise to get 20x^2-6y^2+24 = 0 This is the equivalent of saying that rotating the entire axis system pi/4 radians (or 45 degrees) counter-clockwise will turn 7x^2+26xy+7y^2+24=0 into 20x^2-6y^2+24 = 0

6 years agoAs confirmation, you can graph the two conics and you'll see that one is a rotated version of the other.

6 years agoBut this is an example of what the answer might be...

6 years agoah, so I'm not done yet lol since they want the answer in parametric form

6 years agoone second

6 years agook... i'll look over the first few steps again

6 years agoAre you sure that's the answer? I plotted the answer and I got an ellipse, but the original problem is a hyperbola....so something is off. Is there a typo in the original problem?

6 years agono, that is one of 5 choices... the others are

6 years agooh ok, thank you for providing the other answers

6 years agono problem, you are the one helping me :)

6 years agoIf possible, can I get a screenshot of the original problem?

6 years agosure... oh no! i gave the wrong other answers. hold on

6 years agosure thing

6 years agothe original is Rotate the axes to eliminate the xy-term in the equation. Then write the equation in standard form. 7x^2 + 26xy + 7y^2 - 24 = 0

6 years agothe answers are:

6 years agoah that makes a lot more sense, thx

6 years agoone sec

6 years agoso sorry for the mix up

6 years agono worries

6 years agoStart with the formula given on the wiki page (A*cos^2(theta)+Bsin(theta)*cos(theta)+C*sin^2(theta))*x^2 +(A*sin^2(theta)-Bsin(theta)*cos(theta)+C*cos^2(theta))*y^2 +(D*cos(theta)+E*sin(theta))*x +(-D*sin(theta)+E*cos(theta))*y +F = 0 The formula is broken up into multiple lines (this is one really really big equation). The idea here is to plug in theta = pi/4 and evaluate ------------------------------------------------------- Step 1) Plug in theta = pi/4 (7*cos^2(pi/4)+26*sin(pi/4)*cos(pi/4)+7*sin^2(pi/4))*x^2 + (7*sin^2(pi/4)-26*sin(pi/4)*cos(pi/4)+7*cos^2(pi/4))*y^2 + (0*cos(pi/4)+0*sin(pi/4))*x + (-0*sin(pi/4)+0*cos(pi/4))*y + -24 = 0 ------------------------------------------------------- Step 2) Evaluate the sine and cosine of pi/4 (in this case, they both evaluate to 0.707106781186547 approximately) (7*0.5+26*0.707106781186547*0.707106781186548+7*0.5)*x^2 +(7*0.5-26*0.707106781186547*0.707106781186548+7*0.5)*y^2 +(0*0.707106781186548+0*0.707106781186547)*x +(-0*0.707106781186547+0*0.707106781186548)*y + -24 = 0 ------------------------------------------------------- Step 3) Square each value (7*0.5+26*0.707106781186547*0.707106781186548+7*0.5)*x^2 +(7*0.5-26*0.707106781186547*0.707106781186548+7*0.5)*y^2 +(0*0.707106781186548+0*0.707106781186547)*x +(-0*0.707106781186547+0*0.707106781186548)*y + -24 = 0 ------------------------------------------------------- Step 4) Multiply (3.5+13+3.5)*x^2 +(3.5-13+3.5)*y^2 +(0+0)*x +(0+0)*y + -24 = 0 ------------------------------------------------------- Step 5) Combine like terms (20)*x^2 +(-6)*y^2 +(0)*x +(0)*y + -24 = 0 ------------------------------------------------------- Step 6) Condense into one line (this is now possible since the equation is much shorter/smaller. (20)*x^2+(-6)*y^2+(0)*x+(0)*y+ -24 = 0 ------------------------------------------------------- Step 7) Simplify 20x^2 - 6y^2 - 24 = 0 ------------------------------------------------------- Step 8) Add 24 to both sides 20x^2 - 6y^2 = 24 ------------------------------------------------------- Step 9) Divide every term by 24 20x^2/24 - 6y^2/24 = 24/24 ------------------------------------------------------- Step 10) Simplify and rearrange terms x^2/(6/5) - y^2/4 = 1 ====================================================================================================================== Answer: So the final answer is \[\Large \frac{x^2}{\frac{6}{5}} - \frac{y^2}{4} = 1\] Note: I'm using x in place of x', but they really are the same thing (one is just the converted/rotated coordinate)

6 years agoSo the answer using x' and y' is \[\Large \frac{\left(x^{\prime}\right)^2}{\frac{6}{5}} - \frac{\left(y^{\prime}\right)^2}{4} = 1\]

6 years agoTHANK YOU!!!! no one else was really helpful... i learn much easier by seeing an example of a problem done than just being given a formula.

6 years agoYou're very welcome, I recommend practicing with this idea a lot more because there are a ton of steps which are very very long.

6 years agoi definitely will. i have 4 other similar questions to figure out and really needed step by step instructions.

6 years agooh and I'd go over the wiki page in more depth if you're not sure how the formula works and such

6 years agoThanks

6 years agosure thing

6 years ago