# Thread: rotate coordinate axis &identify the conic.

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

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

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

2. Hello, ramiee2010!

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

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

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

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

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

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

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

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

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

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

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

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

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

Substitute into [1]:

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

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

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

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

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