Menu Close

Question-25002

Tinku Tara 0 Comments
Pin




Question Number 25002 by ajfour last updated on 30/Nov/17
Commented by ajfour last updated on 30/Nov/17
Solution to Q.24966
$${Solution}\:{to}\:{Q}.\mathrm{24966} \\ $$
Answered by ajfour last updated on 30/Nov/17
x^2 −11y^2 −16xy−12x+6y+21=0 Focus =? (x−a)^2 −11(y−b)^2 −16(x−a)(y−b)+c=0 ⇒x^2 −11y^2 −16xy+(−2a+16b)x +(22b+16a)+(a^2 −11b^2 −16ab+c)=0 ⇒ −2a+16b=−12 ...(i) 22b+16a=6 ...(ii) a^2 −11b^2 −16ab+c=21 ...(iii) ⇒ a=(6/5) , b=−(3/5) , c=12 so (x−(6/5))^2 −11(y+(3/5))^2 −16(x−(6/5))(y+(3/5))+12=0 .....(a) let x−(6/5)=Xcos θ−Ysin θ y+(3/5)=Xsin θ+Ycos θ upon substitution in (a): (cos^2 θ−11sin^2 θ−16sin θcos θ)X^2 −(−sin^2 θ+11cos^2 θ−16sin θcos θ)Y^( 2) +(−24sin θcos θ+16sin^2 θ−16cos^2 θ)XY+12=0 If we desire coeff. of XY to be zero ⇒ −24tan θ+16tan^2 θ−16=0 ⇒ tan θ = 2, −(1/2) let us choose tan θ=2 then sin θ=(2/( (√5))) , cos θ=(1/( (√5))) so equation of hyperbola becomes: −15X^2 +5Y^( 2) +12=0 or (X^2 /((((2(√5))/5))^2 ))−(Y^( 2) /((((2(√(15)))/5))^2 )) =1 e^2 =1+(B^2 /A^2 ) =1+(((√(15))/( (√5))))^2 =4 e=2 focus at (x_1 , y_1 ) x_1 =(6/5)+Aecos θ =(6/5)+(((2(√5))/5))×2×(1/( (√5))) = 2 y_1 =−(3/5)+Aesin θ =−(3/5)+(((2(√5))/5))×2×(2/( (√5))) = 1 hence focus at (2, 1) .
$${x}^{\mathrm{2}} −\mathrm{11}{y}^{\mathrm{2}} −\mathrm{16}{xy}−\mathrm{12}{x}+\mathrm{6}{y}+\mathrm{21}=\mathrm{0} \\ $$$${Focus}\:=? \\ $$$$\left({x}−{a}\right)^{\mathrm{2}} −\mathrm{11}\left({y}−{b}\right)^{\mathrm{2}} −\mathrm{16}\left({x}−{a}\right)\left({y}−{b}\right)+{c}=\mathrm{0} \\ $$$$\Rightarrow{x}^{\mathrm{2}} −\mathrm{11}{y}^{\mathrm{2}} −\mathrm{16}{xy}+\left(−\mathrm{2}{a}+\mathrm{16}{b}\right){x} \\ $$$$\:+\left(\mathrm{22}{b}+\mathrm{16}{a}\right)+\left({a}^{\mathrm{2}} −\mathrm{11}{b}^{\mathrm{2}} −\mathrm{16}{ab}+{c}\right)=\mathrm{0} \\ $$$$\Rightarrow\:\:\:\:−\mathrm{2}{a}+\mathrm{16}{b}=−\mathrm{12}\:\:\:\:…\left({i}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{22}{b}+\mathrm{16}{a}=\mathrm{6}\:\:\:\:\:\:\:\:…\left({ii}\right) \\ $$$$\:\:\:{a}^{\mathrm{2}} −\mathrm{11}{b}^{\mathrm{2}} −\mathrm{16}{ab}+{c}=\mathrm{21}\:\:…\left({iii}\right) \\ $$$$\Rightarrow\:\:\:{a}=\frac{\mathrm{6}}{\mathrm{5}}\:,\:\:{b}=−\frac{\mathrm{3}}{\mathrm{5}}\:,\:{c}=\mathrm{12} \\ $$$${so}\:\:\:\:\left({x}−\frac{\mathrm{6}}{\mathrm{5}}\right)^{\mathrm{2}} −\mathrm{11}\left({y}+\frac{\mathrm{3}}{\mathrm{5}}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:−\mathrm{16}\left({x}−\frac{\mathrm{6}}{\mathrm{5}}\right)\left({y}+\frac{\mathrm{3}}{\mathrm{5}}\right)+\mathrm{12}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…..\left({a}\right) \\ $$$${let}\:\:{x}−\frac{\mathrm{6}}{\mathrm{5}}={X}\mathrm{cos}\:\theta−{Y}\mathrm{sin}\:\theta \\ $$$$\:\:\:\:\:\:\:\:{y}+\frac{\mathrm{3}}{\mathrm{5}}={X}\mathrm{sin}\:\theta+{Y}\mathrm{cos}\:\theta \\ $$$${upon}\:{substitution}\:{in}\:\left({a}\right): \\ $$$$\left(\mathrm{cos}\:^{\mathrm{2}} \theta−\mathrm{11sin}\:^{\mathrm{2}} \theta−\mathrm{16sin}\:\theta\mathrm{cos}\:\theta\right){X}^{\mathrm{2}} \\ $$$$\:−\left(−\mathrm{sin}\:^{\mathrm{2}} \theta+\mathrm{11cos}\:^{\mathrm{2}} \theta−\mathrm{16sin}\:\theta\mathrm{cos}\:\theta\right){Y}^{\:\:\mathrm{2}} \\ $$$$+\left(−\mathrm{24sin}\:\theta\mathrm{cos}\:\theta+\mathrm{16sin}\:^{\mathrm{2}} \theta−\mathrm{16cos}\:^{\mathrm{2}} \theta\right){XY}+\mathrm{12}=\mathrm{0} \\ $$$${If}\:{we}\:{desire}\:{coeff}.\:{of}\:{XY}\:{to}\:{be} \\ $$$${zero}\:\:\Rightarrow\: \\ $$$$−\mathrm{24tan}\:\theta+\mathrm{16tan}\:^{\mathrm{2}} \theta−\mathrm{16}=\mathrm{0} \\ $$$$\Rightarrow\:\:\mathrm{tan}\:\theta\:=\:\mathrm{2},\:−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${let}\:{us}\:{choose}\:\mathrm{tan}\:\theta=\mathrm{2} \\ $$$${then}\:\:\mathrm{sin}\:\theta=\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}\:\:\:,\:\:\mathrm{cos}\:\theta=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}} \\ $$$${so}\:{equation}\:{of}\:{hyperbola} \\ $$$${becomes}: \\ $$$$\:\:\:−\mathrm{15}{X}^{\mathrm{2}} +\mathrm{5}{Y}^{\:\:\mathrm{2}} +\mathrm{12}=\mathrm{0} \\ $$$${or}\:\:\:\:\:\:\frac{{X}^{\mathrm{2}} }{\left(\frac{\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{5}}\right)^{\mathrm{2}} }−\frac{{Y}^{\:\:\mathrm{2}} }{\left(\frac{\mathrm{2}\sqrt{\mathrm{15}}}{\mathrm{5}}\right)^{\mathrm{2}} }\:=\mathrm{1} \\ $$$${e}^{\mathrm{2}} =\mathrm{1}+\frac{{B}^{\mathrm{2}} }{{A}^{\mathrm{2}} }\:=\mathrm{1}+\left(\frac{\sqrt{\mathrm{15}}}{\:\sqrt{\mathrm{5}}}\right)^{\mathrm{2}} =\mathrm{4} \\ $$$${e}=\mathrm{2} \\ $$$${focus}\:{at}\:\:\left({x}_{\mathrm{1}} ,\:{y}_{\mathrm{1}} \right) \\ $$$${x}_{\mathrm{1}} =\frac{\mathrm{6}}{\mathrm{5}}+{Ae}\mathrm{cos}\:\theta\: \\ $$$$\:\:\:\:=\frac{\mathrm{6}}{\mathrm{5}}+\left(\frac{\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{5}}\right)×\mathrm{2}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\:=\:\mathrm{2} \\ $$$${y}_{\mathrm{1}} =−\frac{\mathrm{3}}{\mathrm{5}}+{Ae}\mathrm{sin}\:\theta \\ $$$$\:\:\:\:=−\frac{\mathrm{3}}{\mathrm{5}}+\left(\frac{\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{5}}\right)×\mathrm{2}×\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}\:=\:\mathrm{1} \\ $$$${hence}\:{focus}\:{at}\:\:\left(\mathrm{2},\:\mathrm{1}\right)\:. \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *