OpenStudy (anonymous):

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 ago
OpenStudy (asnaseer):

6 years ago
OpenStudy (anonymous):

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

6 years ago
OpenStudy (asnaseer):

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

6 years ago
OpenStudy (asnaseer):

Look at the section titled: Elimination of the xy term by the rotation formula

6 years ago
OpenStudy (anonymous):

But 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 ago
OpenStudy (asnaseer):

which part do you not understand?

6 years ago
OpenStudy (anonymous):

I'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 ago
OpenStudy (asnaseer):

A-C=0 is the correct approach

6 years ago
OpenStudy (asnaseer):

that will lead to:$\tan(2\theta)=\infty$

6 years ago
OpenStudy (asnaseer):

therefore:$2\theta=?$

6 years ago
OpenStudy (anonymous):

∞?

6 years ago
OpenStudy (netlopes1):

if 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 ago
OpenStudy (asnaseer):

which angle will give you infinity when you take its tan?

6 years ago
OpenStudy (anonymous):

i don't know

6 years ago
OpenStudy (asnaseer):

have you studied trigonometry?

6 years ago
OpenStudy (anonymous):

no, this is in my precalc course

6 years ago
OpenStudy (asnaseer):

hmmmm... well you do need to understad basic trigonometry in order to do this question

6 years ago
OpenStudy (anonymous):

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

6 years ago
OpenStudy (asnaseer):

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

6 years ago
OpenStudy (asnaseer):

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

6 years ago
OpenStudy (netlopes1):

Please, how book you use?

6 years ago
OpenStudy (anonymous):

I am using the Larson Hostetler Seventh Edition Precalculus book

6 years ago
OpenStudy (anonymous):

@netlopes1 , how did you get pi/4 ?

6 years ago
OpenStudy (anonymous):

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

6 years ago
OpenStudy (netlopes1):

6 years ago
OpenStudy (anonymous):

yes.. 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 ago
OpenStudy (netlopes1):

Simply, 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 ago
jimthompson5910 (jim_thompson5910):

Ok, 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 ago
jimthompson5910 (jim_thompson5910):

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

6 years ago
OpenStudy (anonymous):

But this is an example of what the answer might be...

6 years ago
jimthompson5910 (jim_thompson5910):

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

6 years ago
jimthompson5910 (jim_thompson5910):

one second

6 years ago
OpenStudy (anonymous):

ok... i'll look over the first few steps again

6 years ago
jimthompson5910 (jim_thompson5910):

Are 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 ago
OpenStudy (anonymous):

no, that is one of 5 choices... the others are

6 years ago
jimthompson5910 (jim_thompson5910):

oh ok, thank you for providing the other answers

6 years ago
OpenStudy (anonymous):

no problem, you are the one helping me :)

6 years ago
jimthompson5910 (jim_thompson5910):

If possible, can I get a screenshot of the original problem?

6 years ago
OpenStudy (anonymous):

sure... oh no! i gave the wrong other answers. hold on

6 years ago
jimthompson5910 (jim_thompson5910):

sure thing

6 years ago
OpenStudy (anonymous):

the 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 ago
OpenStudy (anonymous):

6 years ago
jimthompson5910 (jim_thompson5910):

ah that makes a lot more sense, thx

6 years ago
jimthompson5910 (jim_thompson5910):

one sec

6 years ago
OpenStudy (anonymous):

so sorry for the mix up

6 years ago
jimthompson5910 (jim_thompson5910):

no worries

6 years ago
jimthompson5910 (jim_thompson5910):

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. (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 ago
jimthompson5910 (jim_thompson5910):

So 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 ago
OpenStudy (anonymous):

THANK 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 ago
jimthompson5910 (jim_thompson5910):

You'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 ago
OpenStudy (anonymous):

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

6 years ago
jimthompson5910 (jim_thompson5910):

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

6 years ago
OpenStudy (anonymous):

Thanks

6 years ago
jimthompson5910 (jim_thompson5910):

sure thing

6 years ago