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Thread: rotate coordinate axis &identify the conic.

  1. #1
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    rotate coordinate axis &identify the conic.

    $\displaystyle 17x^2-312xy+108y^2-900=0$

    Raotate the coordinateaxes to remove the xy-term and then identify the type of the conical represent by rhe equation.


    plz hint about Raotate the coordinateaxes to remove the xy-term
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  2. #2
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    Hello, ramiee2010!

    You should be familiar with all the Rotation Formulas
    . . before attempting this type of problem.


    $\displaystyle 17x^2-312xy+108y^2-900\:=\:0$ .[1]

    Rotate the coordinate axes to remove the $\displaystyle xy$-term
    Identify the type of the conic.
    Given: .$\displaystyle Ax^2 + Bxy + Cy^2 + Dx + Ey + F \:=\:0$

    . . The angle of rotation $\displaystyle \theta$ is given by: .$\displaystyle \tan2\theta \:=\:\frac{B}{A-C}$ .[2]

    . . The new coordinates are given by: .$\displaystyle \begin{Bmatrix}x' &=& x\cos\theta - y\sin\theta \\ y' &=& x\sin\theta + y\cos\theta \end{Bmatrix}$ .[3]



    We have: .$\displaystyle A = 17,\;B = -312,\;C = 108$

    Then: .$\displaystyle {\color{blue}[2]}\;\tan2\theta \:=\:\frac{-312}{17-1078} \:=\:\frac{312}{91}$


    $\displaystyle 2\theta$ is in a right triangle with: .$\displaystyle opp = 312,\;adj = 91$
    . . Pythagorus says: .$\displaystyle hyp = 325$

    Hence: .$\displaystyle \sin2\theta = \frac{312}{325},\;\cos2\theta = \frac{91}{325}$



    Identity: .$\displaystyle \tan\theta \:=\:\frac{1-\cos2\theta}{\sin2\theta}$

    We have: .$\displaystyle \tan\theta \;=\;\frac{1-\frac{91}{325}}{\frac{312}{325}} \;=\;\frac{234}{312} \:=\:\frac{3}{4}$

    Hence, $\displaystyle \theta$ is in a 3-4-5 right triangle: .$\displaystyle \sin\theta \,=\,\frac{3}{5},\;\cos\theta \,=\,\frac{4}{5}$


    Substitute into [3]: . $\displaystyle \begin{array}{ccccccccc} x' &=& \frac{4}{5}x - \frac{3}{5}y &=& \dfrac{4x-3y}{5} \\ \\[-3mm] y' &=& \frac{3}{5}x + \frac{4}{5}y &=& \dfrac{3x+4y}{5} \end{array}$


    Substitute into [1]:

    . . $\displaystyle 17\left(\frac{4x-3y}{5}\right)^2 - 312\left(\frac{4x-3y}{5}\right)\left(\frac{3x+4y}{5}\right) + 108\left(\frac{3x+4y}{5}\right)^2 - 900 \;=\;0$

    . . $\displaystyle 17\left(\frac{16x^2 - 24xy + 9y^2}{25}\right) - 312\left(\frac{12x^2 + 7xy - 12y^2}{25}\right) + 108\left(\frac{9x^2 + 24xy + 16y^2}{25}\right)$ .$\displaystyle \;=\;900 $

    . . $\displaystyle 272x^2 - 408xy + 153y^2 - 3744x^2 - 2184xy + 3744y^2 + 972x^2 + 2592xy + 1728y^2 \;=\;25\cdot900 $

    . . . . . . . $\displaystyle -2500x^2 + 5625y^2 \;=\;22,\!500$

    . . . . . . . . . . . $\displaystyle \frac{y^2}{4} - \frac{x^2}{9} \;=\;1 \quad\cdots\quad \text{ hyperbola}$

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