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

This should help you answer this question: http://en.wikipedia.org/wiki/Rotation_of_axes

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

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

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

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.

which part do you not understand?

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

A-C=0 is the correct approach

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

therefore:\[2\theta=?\]

∞?

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 \[\theta=\pi/4\].

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

i don't know

have you studied trigonometry?

no, this is in my precalc course

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

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

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

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

Please, how book you use?

I am using the Larson Hostetler Seventh Edition Precalculus book

@netlopes1 , how did you get pi/4 ?

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

Please, read these pages....

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...

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.

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

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

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

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

one second

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

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?

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

oh ok, thank you for providing the other answers

no problem, you are the one helping me :)

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

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

sure thing

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

the answers are:

ah that makes a lot more sense, thx

one sec

so sorry for the mix up

no worries

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)

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

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.

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.

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

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

Thanks

sure thing

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