Answer to Question #133599 in Analytic Geometry for ryahan

Question #133599
Identify the following conic and hence reduce it to standard form:
36x² + 20xy + 30y² − 70x + 128y + 80 = 0
1
Expert's answer
2020-09-17T13:38:01-0400

The expression for a conic section in the Cartesian coordinate system is defined as:


Ax2+Bxy+Cy2+Dx+Ey+F=0,A0,B0,C0Ax^2+Bxy+Cy^2+Dx+Ey+F=0, A\not=0, B\not=0,C\not=0

The result of B24ACB^2-4AC  determines the type of the conic section obtained.



36x2+20xy+30y270x+128y+80=036x^2+20xy+30y^2-70x+128y+80=018x2+10xy+15y235x+64y+40=018x^2+10xy+15y^2-35x+64y+40=0

A=18,B=10,C=15,D=35,E=64,F=40A=18, B=10, C=15, D=-35, E=64,F=40

B24AC=(10)24(18)(15)=980<0B^2-4AC=(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\theta where θ\theta is defined by cot(2θ)=ACB=181510=0.3=cot2θ1cotθ\cot(2\theta)=\dfrac{A-C}{B}=\dfrac{18-15}{10}=0.3=\dfrac{\cot^2\theta-1}{\cot\theta}



cot2θ0.3cotθ1=0\cot^2\theta-0.3\cot\theta-1=0cotθ=0.3±0.09+42=0.3±4.092\cot\theta=\dfrac{0.3\pm\sqrt{0.09+4}}{2}=\dfrac{0.3\pm\sqrt{4.09}}{2}

Take cotθ=0.3+4.092\cot\theta=\dfrac{0.3+\sqrt{4.09}}{2}

x=xcosθysinθx=x'\cos \theta-y'\sin \theta

y=xsinθ+ycosθy=x'\sin \theta+y'\cos \theta



18(x)2cos2θ36xysinθcosθ+18(y)2sin2θ+18(x')^2\cos^2 \theta-36x'y'\sin\theta\cos\theta+18(y')^2\sin^2\theta++10(x)2sinθcosθ+10xycos2θ10xysin2θ+10(x')^2\sin\theta\cos\theta+10x'y'\cos^2\theta-10x'y'\sin^2\theta-10(y)2sinθcosθ+15(x)2sin2θ+30xysinθcosθ+-10(y')^2\sin\theta\cos\theta+15(x')^2\sin^2\theta+30x'y'\sin\theta\cos\theta++15(y)2cos2θ35xcosθ+35ysinθ++15(y')^2\cos^2\theta-35x'\cos\theta+35y'\sin\theta++64xsinθ+64ycosθ+40=0+64x'\sin \theta+64y'\cos \theta+40=0(x)2(18cos2θ+10sinθcosθ+15sin2θ)+(x')^2(18\cos^2 \theta+10\sin\theta\cos\theta+15\sin^2\theta)++(y)2(18sin2θ10sinθcosθ+15cos2θ)++(y')^2(18\sin^2 \theta-10\sin\theta\cos\theta+15\cos^2\theta)++x(64sinθ35cosθ)+y(64cosθ+35sinθ)+40=0+x'(64\sin\theta-35\cos\theta)+y'(64\cos\theta+35\sin\theta)+40=036sinθcosθ+10cos2θ10sin2θ+30sinθcosθ=-36\sin\theta\cos\theta+10\cos^2\theta-10\sin^2\theta+30\sin\theta\cos\theta==10cos(2θ)3sin(2θ)=10cos(2θ)(13cot(2θ))=0=10\cos(2\theta)-3\sin(2\theta)=10\cos(2\theta)(1-3\cot(2\theta))=0(15+3cos2θ+10sinθcosθ)(x)2+(15+3\cos^2 \theta+10\sin\theta\cos\theta)(x')^2++(15+3sin2θ10sinθcosθ)(y)2++(15+3\sin^2 \theta-10\sin\theta\cos\theta)(y')^2++64xsinθ+64ycosθ+40=0+64x'\sin \theta+64y'\cos \theta+40=0(15+3cos2θ+5sin(2θ))(x+32sinθ15+3cos2θ+5sin(2θ))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+3sin2θ5sin(2θ))(y+32cosθ15+3sin2θ5sin(2θ))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==1024sin2θ15+3cos2θ+5sin(2θ)+=\dfrac{1024\sin^2\theta}{15+3\cos^2 \theta+5\sin(2\theta)}++1024cos2θ15+3sin2θ5sin(2θ)40+\dfrac{1024\cos^2\theta}{15+3\sin^2 \theta-5\sin(2\theta)}-40

u=32sinθ15+3cos2θ+5sin(2θ)u=\dfrac{32\sin\theta}{15+3\cos^2 \theta+5\sin(2\theta)}

v=32cosθ15+3sin2θ5sin(2θ)v=\dfrac{32\cos\theta}{15+3\sin^2 \theta-5\sin(2\theta)}

c2=1024sin2θ15+3cos2θ+5sin(2θ)+c^2=\dfrac{1024\sin^2\theta}{15+3\cos^2 \theta+5\sin(2\theta)}+

+1024cos2θ15+3sin2θ5sin(2θ)40+\dfrac{1024\cos^2\theta}{15+3\sin^2 \theta-5\sin(2\theta)}-40


a2=(15+3cos2θ+5sin(2θ))a^2=(15+3\cos^2 \theta+5\sin(2\theta))

15360+3072sin4θ+3072cos4θ+2560sin(2θ)(15+3cos2θ+5sin(2θ))(15+3sin2θ5sin(2θ))\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))}

b2=15+3sin2θ5sin(2θ)b^2=15+3\sin^2 \theta-5\sin(2\theta)





(x+u)2c2a2+(y+v)2c2b2=1\dfrac{(x'+u)^2}{\dfrac{c^2}{a^2}}+\dfrac{(y'+v)^2}{\dfrac{c^2}{b^2}}=1

Ellipse.



Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

Assignment Expert
17.09.20, 21:18

Dear raihan, You are welcome. We are glad to be helpful. If you liked our service, please press a like-button beside the answer field. Thank you!

raihan
17.09.20, 21:14

thanku

Leave a comment