Answer to Question #133599 in Analytic Geometry for ryahan

Question

Identify the following conic and hence reduce it to standard form:
36x² + 20xy + 30y² − 70x + 128y + 80 = 0

Answer 1

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

$$Ax^2+Bxy+Cy^2+Dx+Ey+F=0, A\not=0, B\not=0,C\not=0$$

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

$$36x^2+20xy+30y^2-70x+128y+80=0$$ $$18x^2+10xy+15y^2-35x+64y+40=0$$

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

$B^2-4AC=(10)^2-4(18)(15)=-980<0$

Then the curve is ellipse, circle, point or no curve.

A. Rotate to remove xy-term

Rotate the coordinate axes counterclockwise through the angle $\theta$ where $\theta$ is defined by $\cot(2\theta)=\dfrac{A-C}{B}=\dfrac{18-15}{10}=0.3=\dfrac{\cot^2\theta-1}{\cot\theta}$

$$\cot^2\theta-0.3\cot\theta-1=0$$ $$\cot\theta=\dfrac{0.3\pm\sqrt{0.09+4}}{2}=\dfrac{0.3\pm\sqrt{4.09}}{2}$$

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

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

$y=x'\sin \theta+y'\cos \theta$

$$18(x')^2\cos^2 \theta-36x'y'\sin\theta\cos\theta+18(y')^2\sin^2\theta+$$ $$+10(x')^2\sin\theta\cos\theta+10x'y'\cos^2\theta-10x'y'\sin^2\theta-$$ $$-10(y')^2\sin\theta\cos\theta+15(x')^2\sin^2\theta+30x'y'\sin\theta\cos\theta+$$ $$+15(y')^2\cos^2\theta-35x'\cos\theta+35y'\sin\theta+$$ $$+64x'\sin \theta+64y'\cos \theta+40=0$$ $$(x')^2(18\cos^2 \theta+10\sin\theta\cos\theta+15\sin^2\theta)+$$ $$+(y')^2(18\sin^2 \theta-10\sin\theta\cos\theta+15\cos^2\theta)+$$ $$+x'(64\sin\theta-35\cos\theta)+y'(64\cos\theta+35\sin\theta)+40=0$$ $$-36\sin\theta\cos\theta+10\cos^2\theta-10\sin^2\theta+30\sin\theta\cos\theta=$$ $$=10\cos(2\theta)-3\sin(2\theta)=10\cos(2\theta)(1-3\cot(2\theta))=0$$ $$(15+3\cos^2 \theta+10\sin\theta\cos\theta)(x')^2+$$ $$+(15+3\sin^2 \theta-10\sin\theta\cos\theta)(y')^2+$$ $$+64x'\sin \theta+64y'\cos \theta+40=0$$ $$(15+3\cos^2 \theta+5\sin(2\theta))(x'+\dfrac{32\sin\theta}{15+3\cos^2 \theta+5\sin(2\theta)})^2+$$ $$+(15+3\sin^2 \theta-5\sin(2\theta))(y'+\dfrac{32\cos\theta}{15+3\sin^2 \theta-5\sin(2\theta)})^2=$$ $$=\dfrac{1024\sin^2\theta}{15+3\cos^2 \theta+5\sin(2\theta)}+$$ $$+\dfrac{1024\cos^2\theta}{15+3\sin^2 \theta-5\sin(2\theta)}-40$$

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

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

$c^2=\dfrac{1024\sin^2\theta}{15+3\cos^2 \theta+5\sin(2\theta)}+$

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

$a^2=(15+3\cos^2 \theta+5\sin(2\theta))$

$\dfrac{15360+3072\sin^4\theta+3072\cos^4\theta+2560\sin(2\theta)}{(15+3\cos^2 \theta+5\sin(2\theta))(15+3\sin^2 \theta-5\sin(2\theta))}$

$b^2=15+3\sin^2 \theta-5\sin(2\theta)$

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

Ellipse.