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Transform-the-equation-5x-2-4xy-2y-2-2x-4y-4-0-into-one-without-xy-x-and-y-terms-




Question Number 26449 by Tinkutara last updated on 25/Dec/17
Transform the equation 5x^2  + 4xy  + 2y^2  − 2x + 4y + 4 = 0 into one  without xy, x and y terms.
$${Transform}\:{the}\:{equation}\:\mathrm{5}{x}^{\mathrm{2}} \:+\:\mathrm{4}{xy} \\ $$$$+\:\mathrm{2}{y}^{\mathrm{2}} \:−\:\mathrm{2}{x}\:+\:\mathrm{4}{y}\:+\:\mathrm{4}\:=\:\mathrm{0}\:{into}\:{one} \\ $$$${without}\:{xy},\:{x}\:{and}\:{y}\:{terms}. \\ $$
Answered by jota@ last updated on 26/Dec/17
  5(x−1)^2 +4(x−1)(y+2)+2(y+2)^2 =1  5u^2 +4uv+2v^2 =1  By rotation tan 2θ=(4/(5−2)) then  sin θ=(1/( (√5)))  and cos θ=(2/( (√5)))  5(((2r−s)/( (√5))))^2 +4(((2r−s)/( (√5))))(((r+2s)/( (√5))))+    +2(((r+2s)/( (√5))))^2 =1  5(4r^2 −4rs+s^2 )+  4(2r^2 +3rs−2s^2 )+  2(r^2 +4rs+4s^2 )=5      30r^2 +5s^2 =5          6r^2 +s^2 =1
$$ \\ $$$$\mathrm{5}\left({x}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{4}\left({x}−\mathrm{1}\right)\left({y}+\mathrm{2}\right)+\mathrm{2}\left({y}+\mathrm{2}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$$\mathrm{5}{u}^{\mathrm{2}} +\mathrm{4}{uv}+\mathrm{2}{v}^{\mathrm{2}} =\mathrm{1} \\ $$$${By}\:{rotation}\:\mathrm{tan}\:\mathrm{2}\theta=\frac{\mathrm{4}}{\mathrm{5}−\mathrm{2}}\:{then} \\ $$$$\mathrm{sin}\:\theta=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\:\:{and}\:\mathrm{cos}\:\theta=\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}} \\ $$$$\mathrm{5}\left(\frac{\mathrm{2}{r}−{s}}{\:\sqrt{\mathrm{5}}}\right)^{\mathrm{2}} +\mathrm{4}\left(\frac{\mathrm{2}{r}−{s}}{\:\sqrt{\mathrm{5}}}\right)\left(\frac{{r}+\mathrm{2}{s}}{\:\sqrt{\mathrm{5}}}\right)+ \\ $$$$\:\:+\mathrm{2}\left(\frac{{r}+\mathrm{2}{s}}{\:\sqrt{\mathrm{5}}}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$$\mathrm{5}\left(\mathrm{4}{r}^{\mathrm{2}} −\mathrm{4}{rs}+{s}^{\mathrm{2}} \right)+ \\ $$$$\mathrm{4}\left(\mathrm{2}{r}^{\mathrm{2}} +\mathrm{3}{rs}−\mathrm{2}{s}^{\mathrm{2}} \right)+ \\ $$$$\mathrm{2}\left({r}^{\mathrm{2}} +\mathrm{4}{rs}+\mathrm{4}{s}^{\mathrm{2}} \right)=\mathrm{5} \\ $$$$\:\:\:\:\mathrm{30}{r}^{\mathrm{2}} +\mathrm{5}{s}^{\mathrm{2}} =\mathrm{5}\: \\ $$$$\:\:\:\:\:\:\:\mathrm{6}{r}^{\mathrm{2}} +{s}^{\mathrm{2}} =\mathrm{1} \\ $$$$\:\:\:\:\:\:\: \\ $$
Commented by Tinkutara last updated on 27/Dec/17
Thanks for this but how rotation formula is applied? Isn't it for angle bisectors between pairs of lines passing through origin?
Commented by ajfour last updated on 27/Dec/17
angular bisectors of a pair of straight lines is the locus of points having same distance from the two lines. Rotation wont be helpful.
Commented by Tinkutara last updated on 27/Dec/17
How you replaced u and v from 2nd  line after finding θ?
$${How}\:{you}\:{replaced}\:{u}\:{and}\:{v}\:{from}\:\mathrm{2}{nd} \\ $$$${line}\:{after}\:{finding}\:\theta? \\ $$
Commented by ajfour last updated on 27/Dec/17
Commented by Tinkutara last updated on 27/Dec/17
I got it. Rotation will eliminate xy terms.
Commented by jota@ last updated on 28/Dec/17
find a and b  u=x−a and  v=y−b  with  5u^2 +4uv+2v^2 +k≡5x^2 +4xy+     +2y^2 −2x+4y+4.
$${find}\:{a}\:{and}\:{b} \\ $$$${u}={x}−{a}\:{and}\:\:{v}={y}−{b}\:\:{with} \\ $$$$\mathrm{5}{u}^{\mathrm{2}} +\mathrm{4}{uv}+\mathrm{2}{v}^{\mathrm{2}} +{k}\equiv\mathrm{5}{x}^{\mathrm{2}} +\mathrm{4}{xy}+ \\ $$$$\:\:\:+\mathrm{2}{y}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{4}{y}+\mathrm{4}. \\ $$$$ \\ $$

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