# rotate coordinate axis &identify the conic.

• Jun 6th 2010, 08:07 AM
ramiee2010
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
• Jun 6th 2010, 10:17 AM
Soroban
Hello, ramiee2010!

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

Quote:

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

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

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

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

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

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