The expression for a conic section in the Cartesian coordinate system is defined as:
A x 2 + B x y + C y 2 + D x + E y + F = 0 , A ≠ 0 , B ≠ 0 , C ≠ 0 Ax^2+Bxy+Cy^2+Dx+Ey+F=0, A\not=0, B\not=0,C\not=0 A x 2 + B x y + C y 2 + D x + E y + F = 0 , A = 0 , B = 0 , C = 0 The result of B 2 − 4 A C B^2-4AC B 2 − 4 A C
36 x 2 + 20 x y + 30 y 2 − 70 x + 128 y + 80 = 0 36x^2+20xy+30y^2-70x+128y+80=0 36 x 2 + 20 x y + 30 y 2 − 70 x + 128 y + 80 = 0 18 x 2 + 10 x y + 15 y 2 − 35 x + 64 y + 40 = 0 18x^2+10xy+15y^2-35x+64y+40=0 18 x 2 + 10 x y + 15 y 2 − 35 x + 64 y + 40 = 0 A = 18 , B = 10 , C = 15 , D = − 35 , E = 64 , F = 40 A=18, B=10, C=15, D=-35, E=64,F=40 A = 18 , B = 10 , C = 15 , D = − 35 , E = 64 , F = 40
B 2 − 4 A C = ( 10 ) 2 − 4 ( 18 ) ( 15 ) = − 980 < 0 B^2-4AC=(10)^2-4(18)(15)=-980<0 B 2 − 4 A C = ( 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 \theta θ θ \theta θ cot ( 2 θ ) = A − C B = 18 − 15 10 = 0.3 = cot 2 θ − 1 cot θ \cot(2\theta)=\dfrac{A-C}{B}=\dfrac{18-15}{10}=0.3=\dfrac{\cot^2\theta-1}{\cot\theta} cot ( 2 θ ) = B A − C = 10 18 − 15 = 0.3 = cot θ cot 2 θ − 1
cot 2 θ − 0.3 cot θ − 1 = 0 \cot^2\theta-0.3\cot\theta-1=0 cot 2 θ − 0.3 cot θ − 1 = 0 cot θ = 0.3 ± 0.09 + 4 2 = 0.3 ± 4.09 2 \cot\theta=\dfrac{0.3\pm\sqrt{0.09+4}}{2}=\dfrac{0.3\pm\sqrt{4.09}}{2} cot θ = 2 0.3 ± 0.09 + 4 = 2 0.3 ± 4.09 Take cot θ = 0.3 + 4.09 2 \cot\theta=\dfrac{0.3+\sqrt{4.09}}{2} cot θ = 2 0.3 + 4.09
x = x ′ cos θ − y ′ sin θ x=x'\cos \theta-y'\sin \theta x = x ′ cos θ − y ′ sin θ
y = x ′ sin θ + y ′ cos θ y=x'\sin \theta+y'\cos \theta y = x ′ sin θ + y ′ cos θ
18 ( x ′ ) 2 cos 2 θ − 36 x ′ y ′ sin θ cos θ + 18 ( y ′ ) 2 sin 2 θ + 18(x')^2\cos^2 \theta-36x'y'\sin\theta\cos\theta+18(y')^2\sin^2\theta+ 18 ( x ′ ) 2 cos 2 θ − 36 x ′ y ′ sin θ cos θ + 18 ( y ′ ) 2 sin 2 θ + + 10 ( x ′ ) 2 sin θ cos θ + 10 x ′ y ′ cos 2 θ − 10 x ′ y ′ sin 2 θ − +10(x')^2\sin\theta\cos\theta+10x'y'\cos^2\theta-10x'y'\sin^2\theta- + 10 ( x ′ ) 2 sin θ cos θ + 10 x ′ y ′ cos 2 θ − 10 x ′ y ′ sin 2 θ − − 10 ( y ′ ) 2 sin θ cos θ + 15 ( x ′ ) 2 sin 2 θ + 30 x ′ y ′ sin θ cos θ + -10(y')^2\sin\theta\cos\theta+15(x')^2\sin^2\theta+30x'y'\sin\theta\cos\theta+ − 10 ( y ′ ) 2 sin θ cos θ + 15 ( x ′ ) 2 sin 2 θ + 30 x ′ y ′ sin θ cos θ + + 15 ( y ′ ) 2 cos 2 θ − 35 x ′ cos θ + 35 y ′ sin θ + +15(y')^2\cos^2\theta-35x'\cos\theta+35y'\sin\theta+ + 15 ( y ′ ) 2 cos 2 θ − 35 x ′ cos θ + 35 y ′ sin θ + + 64 x ′ sin θ + 64 y ′ cos θ + 40 = 0 +64x'\sin \theta+64y'\cos \theta+40=0 + 64 x ′ sin θ + 64 y ′ cos θ + 40 = 0 ( x ′ ) 2 ( 18 cos 2 θ + 10 sin θ cos θ + 15 sin 2 θ ) + (x')^2(18\cos^2 \theta+10\sin\theta\cos\theta+15\sin^2\theta)+ ( x ′ ) 2 ( 18 cos 2 θ + 10 sin θ cos θ + 15 sin 2 θ ) + + ( y ′ ) 2 ( 18 sin 2 θ − 10 sin θ cos θ + 15 cos 2 θ ) + +(y')^2(18\sin^2 \theta-10\sin\theta\cos\theta+15\cos^2\theta)+ + ( y ′ ) 2 ( 18 sin 2 θ − 10 sin θ cos θ + 15 cos 2 θ ) + + x ′ ( 64 sin θ − 35 cos θ ) + y ′ ( 64 cos θ + 35 sin θ ) + 40 = 0 +x'(64\sin\theta-35\cos\theta)+y'(64\cos\theta+35\sin\theta)+40=0 + x ′ ( 64 sin θ − 35 cos θ ) + y ′ ( 64 cos θ + 35 sin θ ) + 40 = 0 − 36 sin θ cos θ + 10 cos 2 θ − 10 sin 2 θ + 30 sin θ cos θ = -36\sin\theta\cos\theta+10\cos^2\theta-10\sin^2\theta+30\sin\theta\cos\theta= − 36 sin θ cos θ + 10 cos 2 θ − 10 sin 2 θ + 30 sin θ cos θ = = 10 cos ( 2 θ ) − 3 sin ( 2 θ ) = 10 cos ( 2 θ ) ( 1 − 3 cot ( 2 θ ) ) = 0 =10\cos(2\theta)-3\sin(2\theta)=10\cos(2\theta)(1-3\cot(2\theta))=0 = 10 cos ( 2 θ ) − 3 sin ( 2 θ ) = 10 cos ( 2 θ ) ( 1 − 3 cot ( 2 θ )) = 0 ( 15 + 3 cos 2 θ + 10 sin θ cos θ ) ( x ′ ) 2 + (15+3\cos^2 \theta+10\sin\theta\cos\theta)(x')^2+ ( 15 + 3 cos 2 θ + 10 sin θ cos θ ) ( x ′ ) 2 + + ( 15 + 3 sin 2 θ − 10 sin θ cos θ ) ( y ′ ) 2 + +(15+3\sin^2 \theta-10\sin\theta\cos\theta)(y')^2+ + ( 15 + 3 sin 2 θ − 10 sin θ cos θ ) ( y ′ ) 2 + + 64 x ′ sin θ + 64 y ′ cos θ + 40 = 0 +64x'\sin \theta+64y'\cos \theta+40=0 + 64 x ′ sin θ + 64 y ′ cos θ + 40 = 0 ( 15 + 3 cos 2 θ + 5 sin ( 2 θ ) ) ( x ′ + 32 sin θ 15 + 3 cos 2 θ + 5 sin ( 2 θ ) ) 2 + (15+3\cos^2 \theta+5\sin(2\theta))(x'+\dfrac{32\sin\theta}{15+3\cos^2 \theta+5\sin(2\theta)})^2+ ( 15 + 3 cos 2 θ + 5 sin ( 2 θ )) ( x ′ + 15 + 3 cos 2 θ + 5 sin ( 2 θ ) 32 sin θ ) 2 + + ( 15 + 3 sin 2 θ − 5 sin ( 2 θ ) ) ( y ′ + 32 cos θ 15 + 3 sin 2 θ − 5 sin ( 2 θ ) ) 2 = +(15+3\sin^2 \theta-5\sin(2\theta))(y'+\dfrac{32\cos\theta}{15+3\sin^2 \theta-5\sin(2\theta)})^2= + ( 15 + 3 sin 2 θ − 5 sin ( 2 θ )) ( y ′ + 15 + 3 sin 2 θ − 5 sin ( 2 θ ) 32 cos θ ) 2 = = 1024 sin 2 θ 15 + 3 cos 2 θ + 5 sin ( 2 θ ) + =\dfrac{1024\sin^2\theta}{15+3\cos^2 \theta+5\sin(2\theta)}+ = 15 + 3 cos 2 θ + 5 sin ( 2 θ ) 1024 sin 2 θ + + 1024 cos 2 θ 15 + 3 sin 2 θ − 5 sin ( 2 θ ) − 40 +\dfrac{1024\cos^2\theta}{15+3\sin^2 \theta-5\sin(2\theta)}-40 + 15 + 3 sin 2 θ − 5 sin ( 2 θ ) 1024 cos 2 θ − 40 u = 32 sin θ 15 + 3 cos 2 θ + 5 sin ( 2 θ ) u=\dfrac{32\sin\theta}{15+3\cos^2 \theta+5\sin(2\theta)} u = 15 + 3 cos 2 θ + 5 sin ( 2 θ ) 32 sin θ
v = 32 cos θ 15 + 3 sin 2 θ − 5 sin ( 2 θ ) v=\dfrac{32\cos\theta}{15+3\sin^2 \theta-5\sin(2\theta)} v = 15 + 3 sin 2 θ − 5 sin ( 2 θ ) 32 cos θ
c 2 = 1024 sin 2 θ 15 + 3 cos 2 θ + 5 sin ( 2 θ ) + c^2=\dfrac{1024\sin^2\theta}{15+3\cos^2 \theta+5\sin(2\theta)}+ c 2 = 15 + 3 cos 2 θ + 5 sin ( 2 θ ) 1024 sin 2 θ +
+ 1024 cos 2 θ 15 + 3 sin 2 θ − 5 sin ( 2 θ ) − 40 +\dfrac{1024\cos^2\theta}{15+3\sin^2 \theta-5\sin(2\theta)}-40 + 15 + 3 sin 2 θ − 5 sin ( 2 θ ) 1024 cos 2 θ − 40
a 2 = ( 15 + 3 cos 2 θ + 5 sin ( 2 θ ) ) a^2=(15+3\cos^2 \theta+5\sin(2\theta)) a 2 = ( 15 + 3 cos 2 θ + 5 sin ( 2 θ ))
15360 + 3072 sin 4 θ + 3072 cos 4 θ + 2560 sin ( 2 θ ) ( 15 + 3 cos 2 θ + 5 sin ( 2 θ ) ) ( 15 + 3 sin 2 θ − 5 sin ( 2 θ ) ) \dfrac{15360+3072\sin^4\theta+3072\cos^4\theta+2560\sin(2\theta)}{(15+3\cos^2 \theta+5\sin(2\theta))(15+3\sin^2 \theta-5\sin(2\theta))} ( 15 + 3 cos 2 θ + 5 sin ( 2 θ )) ( 15 + 3 sin 2 θ − 5 sin ( 2 θ )) 15360 + 3072 sin 4 θ + 3072 cos 4 θ + 2560 sin ( 2 θ )
b 2 = 15 + 3 sin 2 θ − 5 sin ( 2 θ ) b^2=15+3\sin^2 \theta-5\sin(2\theta) b 2 = 15 + 3 sin 2 θ − 5 sin ( 2 θ )
( x ′ + u ) 2 c 2 a 2 + ( y ′ + v ) 2 c 2 b 2 = 1 \dfrac{(x'+u)^2}{\dfrac{c^2}{a^2}}+\dfrac{(y'+v)^2}{\dfrac{c^2}{b^2}}=1 a 2 c 2 ( x ′ + u ) 2 + b 2 c 2 ( y ′ + v ) 2 = 1 Ellipse.
#340153 on Dec 2023
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