Question

question 2 (5 points)
listen
use \\( \theta=45^{circ} \\) to write \\( 2 x^{2}+2 x y+5 y^{2}+6 x-4 y=40 \\) in the \\( x^{prime} y^{prime} \\)-plaric. then identify the conic.
\\( \left(172 \sqrt{3}\
ight)\left(x^{\prime}\
ight)^{2}+(-46 \sqrt{3}) x^{\prime} y^{\prime}+(+11-2 \sqrt{3})\left(y^{\prime}\
ight)^{2}+(12 \sqrt{2}-8 \sqrt{3}) x^{\prime}+(-8 \sqrt{2}-12 \sqrt{3}) y^{\prime}-160=0 \\); parabola
\\( \left(172 \sqrt{3}\
ight)\left(x^{\prime}\
ight)^{2}+(-46 \sqrt{3}) x^{\prime} y^{\prime}+(+11-2 \sqrt{3})\left(y^{\prime}\
ight)^{2}+(12 \sqrt{2}-8 \sqrt{3}) x^{\prime}+(-8 \sqrt{2}-12 \sqrt{3}) y^{\prime}-160=0 \\); cllipse
\\( (9)\left(x^{\prime}\
ight)^{2}+(6) x^{\prime} y^{\prime}+(5)\left(y^{\prime}\
ight)^{2}+(10 \sqrt{2}) x^{\prime}+(-10 \sqrt{2}) y^{\prime}-80=0 \\); parabola
\\( (9)\left(x^{\prime}\
ight)^{2}+(6) x^{\prime} y^{\prime}+(5)\left(y^{\prime}\
ight)^{2}+(10 \sqrt{2}) x^{\prime}+(-10 \sqrt{2}) y^{\prime}-80=0 \\); ellipse

Explanation:

Step1: Recall the rotation of axes formulas

The rotation of axes formulas are \(x = x'\cos\theta - y'\sin\theta\) and \(y=x'\sin\theta + y'\cos\theta\). Since \(\theta = 45^{\circ}\), \(\cos\theta=\sin\theta=\frac{\sqrt{2}}{2}\). So \(x=\frac{\sqrt{2}}{2}(x' - y')\) and \(y=\frac{\sqrt{2}}{2}(x'+y')\).

Step2: Substitute \(x\) and \(y\) into the given equation

Substitute \(x\) and \(y\) into \(2x^{2}+2xy + 5y^{2}+6x-4y = 40\).
First, \(x^{2}=2x'^{2}-2x'y'+y'^{2}\), \(xy=\frac{1}{2}(x'^{2}-y'^{2})\), \(y^{2}=\frac{1}{2}(x'^{2}+2x'y'+y'^{2})\), \(x=\frac{\sqrt{2}}{2}(x' - y')\), \(y=\frac{\sqrt{2}}{2}(x'+y')\).

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LATEXBLOCK0
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For a general second - degree equation \(Ax'^{2}+Bx'y'+Cy'^{2}+Dx'+Ey'+F = 0\), the discriminant is \(B^{2}-4AC\). Here \(A = 9\), \(B = 6\), \(C = 5\), so \(B^{2}-4AC=36-180=- 144<0\). When \(B^{2}-4AC<0\), the conic is an ellipse.

Answer:

\((9)(x')^{2}+(6)x'y'+(5)(y')^{2}+(10\sqrt{2})x'+(-10\sqrt{2})y'-80 = 0\); ellipse