Question #155716

ansform the equation 17π‘₯

2 + 18π‘₯𝑦 βˆ’ 7𝑦

2 βˆ’ 16π‘₯ βˆ’ 32𝑦 βˆ’ 18 = 0 to one in 

which there is no term involving π‘₯, 𝑦.


1
Expert's answer
2021-01-20T17:47:19-0500

17π‘₯2+18π‘₯π‘¦βˆ’7𝑦2βˆ’16π‘₯βˆ’32π‘¦βˆ’18=017π‘₯^2 + 18π‘₯𝑦 βˆ’ 7𝑦^2 βˆ’ 16π‘₯ βˆ’ 32𝑦 βˆ’ 18 = 0


The rotation angle:

tan2ΞΈ=BAβˆ’Ctan2\theta=\frac{B}{A-C}

where A, B and C are the corresponding coefficients from the equation:

Ax2+Bxy+Cy2+Dx+Ey+F=0Ax^2+Bxy+Cy^2+Dx+Ey+F=0

tan2ΞΈ=181+7=34=2tanΞΈ1βˆ’tan2ΞΈtan2\theta=\frac{18}{1+7}=\frac{3}{4}=\frac{2tan\theta}{1-tan^2\theta}

tanΞΈ=13tan\theta=\frac{1}{3}


The required transformation equations needed to rotate the coordinate axes to eliminate the xy

term in the given equation for the conic section are:

x=xβ€²cosΞΈβˆ’yβ€²sinΞΈx=x'cos\theta-y'sin\theta

y=xβ€²cosΞΈ+yβ€²sinΞΈy=x'cos\theta+y'sin\theta

cosΞΈ=3/10,sinΞΈ=1/10cos\theta=3/\sqrt{10}, sin\theta=1/\sqrt{10}

x=3xβ€²βˆ’yβ€²10,y=xβ€²+3yβ€²10x=\frac{3x'-y'}{\sqrt{10}}, y=\frac{x'+3y'}{\sqrt{10}}


Substituting these expressions for x and y into the original equation for the conic section and simplifying, we get:

20xβ€²2βˆ’10yβ€²2βˆ’8010xβ€²βˆ’8010yβ€²βˆ’18=020x^{'2}-10y^{'2}-\frac{80}{\sqrt{10}}x'-\frac{80}{\sqrt{10}}y'-18=0


Simplifying and completing the squares to obtain an equation for a hyperbola in standard form, we get:


(xβ€²βˆ’2/10)21/2βˆ’(yβ€²+4/10)21=1\frac{(x'-2/\sqrt{10})^2}{1/2}-\frac{(y'+4/\sqrt{10})^2}{1}=1


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