Results 1 to 3 of 3

Math Help - Rotation of Conic

  1. #1
    Newbie
    Joined
    Nov 2007
    Posts
    4

    Exclamation Rotation of Conic

    Hi,

    Please help me with this problem. I tried to solve it but I got stuck ...

    Problem: Rotate the axes so that the new equation contains no xy-term. Discuss the new equation.

    x^2 + 2xy + y^2 - 8x + 8y = 0

    Here is what I got so far:

    Solution:

    A = 1, B = 2, C =1

    cot 2 theta = (A - C) / B ==> (1 - 1)/2 ==> 0/2
    (cot^2 theta - 1)/ 2 cot theta = 0/2
    2 cot^2 theta - 2 = 0
    2(cot^2 theta - 1) = 0
    2(cot theta - 1) (cot theta + 1) = 0
    cot theta = 1
    theta = 45 degree

    x = x' cos theta - y' sin theta
    x = x' cos 45 - y' sin 45

    y = x' sin 45 + y' cos 45

    i got to this far and plug in to the original equation and simplied it. But I don't know what to do after that. Please help me!

    Thank you!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,660
    Thanks
    600
    Hello, fallingsky!

    Rotate the axes so that the new equation contains no xy-term.
    . . <br />
x^2 + 2xy + y^2 - 8x + 8y \;= \;0

    Here is what I got so far: . A = 1,\;B = 2,\;C =1

    \cot 2\theta \:= \:\frac{A - C}{B} \:=\:\frac{0}{2} \:=\:0

    \cot2\theta \:=\:0\quad\Rightarrow\quad 2\theta \:=\:90^o \quad\Rightarrow\quad \theta \:=\:45^o


    x \:= \:x'\cos45^o - y'\sin45^o \:=\:\frac{1}{\sqrt{2}}(x - y)

    y \:= \:x'\sin45^o + y'\cos45^o \;=\;\frac{1}{\sqrt{2}}(x + y) . . . . all correct so far!

    i got to this far and plug in to the original equation and simplied it.
    But I don't know what to do after that.
    Do you recognize the equation?

    Substitute:
    . \left[\frac{1}{\sqrt{2}}(x-y)\right]^2 + 2\left[\frac{1}{\sqrt{2}}(x-y)\right]\left[\frac{1}{\sqrt{2}}(x+y)\right] + \left[\frac{1}{\sqrt{2}}(x + y)\right]^2  -8\left[\frac{1}{\sqrt{2}}(x-y)\right] + 8\left[\frac{1}{\sqrt{2}}(x + y)\right] \;=\;0

    . . . . \frac{1}{2}x^2 - xy + \frac{1}{2}y^2 + x^2-y^2 + \frac{1}{2}x^2 + xy + \frac{1}{2}y^2 <br />
-4\sqrt{2} + 4\sqrt{2}y + 4\sqrt{2}x + 4\sqrt{2}y\;=\;0

    . . . . x^2 + 4\sqrt{2}y \;=\;0\quad\Rightarrow\quad x^2 \:=\:-4\sqrt{2}y


    This is a parabola, opening upward, vertex at the origin.

    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Nov 2007
    Posts
    4
    Thank you so much Soroban! I got confused at the end when I tried to substitute it in the original problem, so I stopped and asked for help. .. I should have think more about this before asking for help. Again, thank you so much sir! Happy Thanksgiving!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 0
    Last Post: January 14th 2012, 08:02 PM
  2. Rotation of a conic?
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: May 20th 2011, 12:20 AM
  3. Conic Rotation
    Posted in the Geometry Forum
    Replies: 2
    Last Post: November 14th 2009, 05:39 AM
  4. Conic Second
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: September 16th 2008, 12:01 AM
  5. hyperbola and conic rotation
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: April 29th 2007, 11:23 AM

Search Tags


/mathhelpforum @mathhelpforum