The expression for a conic section in the Cartesian coordinate system is defined as:
Ax2+Bxy+Cy2+Dx+Ey+F=0,A=0,B=0,C=0The result of B2−4AC determines the type of the conic section obtained.
36x2+20xy+30y2−70x+128y+80=018x2+10xy+15y2−35x+64y+40=0A=18,B=10,C=15,D=−35,E=64,F=40
B2−4AC=(10)2−4(18)(15)=−980<0
Then the curve is ellipse, circle, point or no curve.
A. Rotate to remove xy-term
Rotate the coordinate axes counterclockwise through the angle θ where θ is defined by cot(2θ)=BA−C=1018−15=0.3=cotθcot2θ−1
cot2θ−0.3cotθ−1=0cotθ=20.3±0.09+4=20.3±4.09Take cotθ=20.3+4.09
x=x′cosθ−y′sinθ
y=x′sinθ+y′cosθ
18(x′)2cos2θ−36x′y′sinθcosθ+18(y′)2sin2θ++10(x′)2sinθcosθ+10x′y′cos2θ−10x′y′sin2θ−−10(y′)2sinθcosθ+15(x′)2sin2θ+30x′y′sinθcosθ++15(y′)2cos2θ−35x′cosθ+35y′sinθ++64x′sinθ+64y′cosθ+40=0(x′)2(18cos2θ+10sinθcosθ+15sin2θ)++(y′)2(18sin2θ−10sinθcosθ+15cos2θ)++x′(64sinθ−35cosθ)+y′(64cosθ+35sinθ)+40=0−36sinθcosθ+10cos2θ−10sin2θ+30sinθcosθ==10cos(2θ)−3sin(2θ)=10cos(2θ)(1−3cot(2θ))=0(15+3cos2θ+10sinθcosθ)(x′)2++(15+3sin2θ−10sinθcosθ)(y′)2++64x′sinθ+64y′cosθ+40=0(15+3cos2θ+5sin(2θ))(x′+15+3cos2θ+5sin(2θ)32sinθ)2++(15+3sin2θ−5sin(2θ))(y′+15+3sin2θ−5sin(2θ)32cosθ)2==15+3cos2θ+5sin(2θ)1024sin2θ++15+3sin2θ−5sin(2θ)1024cos2θ−40u=15+3cos2θ+5sin(2θ)32sinθ
v=15+3sin2θ−5sin(2θ)32cosθ
c2=15+3cos2θ+5sin(2θ)1024sin2θ+
+15+3sin2θ−5sin(2θ)1024cos2θ−40
a2=(15+3cos2θ+5sin(2θ))
(15+3cos2θ+5sin(2θ))(15+3sin2θ−5sin(2θ))15360+3072sin4θ+3072cos4θ+2560sin(2θ)
b2=15+3sin2θ−5sin(2θ)
a2c2(x′+u)2+b2c2(y′+v)2=1Ellipse.
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