Menu Close

Question-221184

Tinku Tara 0 Comments
Pin




Question Number 221184 by Lekhraj last updated on 26/May/25
Commented by mehdee7396 last updated on 26/May/25
A,B ? are thay focies
$${A},{B}\:?\:{are}\:{thay}\:{focies} \\ $$
Commented by Lekhraj last updated on 26/May/25
No
$$\mathrm{No} \\ $$
Answered by vnm last updated on 26/May/25
((2000)/(77))
$$\frac{\mathrm{2000}}{\mathrm{77}} \\ $$
Commented by Lekhraj last updated on 26/May/25
How?
$$\mathrm{How}? \\ $$
Commented by vnm last updated on 26/May/25
AB=q ∡OBA=θ B=(qcosθ, 0) A=(0, qsinθ) C=(q(cosθ+sinθ), qcosθ) D=(qsinθ, q(sinθ+cosθ)) { ((((q^2 (cosθ+sinθ)^2 )/(80))+((q^2 cos^2 θ)/(50))=1)),((((q^2 sin^2 θ)/(80))+((q^2 (sinθ+cosθ)^2 )/(50))=1)) :} q^2 (((1+sin2θ)/(80))+((1+cos2θ)/(100)))=1 5+5sin2θ+4+4cos2θ=((400)/q^2 ) 9+5((2tgθ)/(1+tg^2 θ))+4((1−tg^2 θ)/(1+tg^2 θ))=((400)/q^2 ) (5−((400)/q^2 ))tg^2 θ+10tgθ+(13−((400)/q^2 ))=0 tgθ=((−10−(√(100−4(5−((400)/q^2 ))(13−((400)/q^2 )))))/(2(5−((400)/q^2 )))) ((q^2 (1−cos2θ))/(160))+((q^2 (1+sin2θ))/(50))=1 5(1−cos2θ)+16(1+sin2θ)=((800)/q^2 ) 21−5cos2θ+16sin2θ=((800)/q^2 ) 21−5((1−tg^2 θ)/(1+tg^2 θ))+16((2tgθ)/(1+tg^2 θ))=((800)/q^2 ) (26−((800)/q^2 ))tg^2 θ+32tgθ+(16−((800)/q^2 ))=0 tgθ=((−32+(√(1024−4(26−((800)/q^2 ))(16−((800)/q^2 )))))/(2(26−((800)/q^2 )))) t=((400)/q^2 ) ((−16+2(√(−40+21t−t^2 )))/(13−t))=((−10−2(√(−40+18t−t^2 )))/(5−t)) t=((77)/5) S_(ABCD) =q^2 =((400)/t)=((2000)/(77))
$$ \\ $$$${AB}={q} \\ $$$$\measuredangle{OBA}=\theta \\ $$$${B}=\left({q}\mathrm{cos}\theta,\:\mathrm{0}\right) \\ $$$${A}=\left(\mathrm{0},\:{q}\mathrm{sin}\theta\right) \\ $$$${C}=\left({q}\left(\mathrm{cos}\theta+\mathrm{sin}\theta\right),\:{q}\mathrm{cos}\theta\right) \\ $$$${D}=\left({q}\mathrm{sin}\theta,\:{q}\left(\mathrm{sin}\theta+\mathrm{cos}\theta\right)\right) \\ $$$$\begin{cases}{\frac{{q}^{\mathrm{2}} \left(\mathrm{cos}\theta+\mathrm{sin}\theta\right)^{\mathrm{2}} }{\mathrm{80}}+\frac{{q}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \theta}{\mathrm{50}}=\mathrm{1}}\\{\frac{{q}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \theta}{\mathrm{80}}+\frac{{q}^{\mathrm{2}} \left(\mathrm{sin}\theta+\mathrm{cos}\theta\right)^{\mathrm{2}} }{\mathrm{50}}=\mathrm{1}}\end{cases} \\ $$$${q}^{\mathrm{2}} \left(\frac{\mathrm{1}+\mathrm{sin2}\theta}{\mathrm{80}}+\frac{\mathrm{1}+\mathrm{cos2}\theta}{\mathrm{100}}\right)=\mathrm{1} \\ $$$$\mathrm{5}+\mathrm{5sin2}\theta+\mathrm{4}+\mathrm{4cos2}\theta=\frac{\mathrm{400}}{{q}^{\mathrm{2}} } \\ $$$$\mathrm{9}+\mathrm{5}\frac{\mathrm{2tg}\theta}{\mathrm{1}+\mathrm{tg}^{\mathrm{2}} \theta}+\mathrm{4}\frac{\mathrm{1}−\mathrm{tg}^{\mathrm{2}} \theta}{\mathrm{1}+\mathrm{tg}^{\mathrm{2}} \theta}=\frac{\mathrm{400}}{{q}^{\mathrm{2}} } \\ $$$$\left(\mathrm{5}−\frac{\mathrm{400}}{{q}^{\mathrm{2}} }\right)\mathrm{tg}^{\mathrm{2}} \theta+\mathrm{10tg}\theta+\left(\mathrm{13}−\frac{\mathrm{400}}{{q}^{\mathrm{2}} }\right)=\mathrm{0} \\ $$$$\mathrm{tg}\theta=\frac{−\mathrm{10}−\sqrt{\mathrm{100}−\mathrm{4}\left(\mathrm{5}−\frac{\mathrm{400}}{{q}^{\mathrm{2}} }\right)\left(\mathrm{13}−\frac{\mathrm{400}}{{q}^{\mathrm{2}} }\right)}}{\mathrm{2}\left(\mathrm{5}−\frac{\mathrm{400}}{{q}^{\mathrm{2}} }\right)} \\ $$$$\frac{{q}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{cos2}\theta\right)}{\mathrm{160}}+\frac{{q}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{sin2}\theta\right)}{\mathrm{50}}=\mathrm{1} \\ $$$$\mathrm{5}\left(\mathrm{1}−\mathrm{cos2}\theta\right)+\mathrm{16}\left(\mathrm{1}+\mathrm{sin2}\theta\right)=\frac{\mathrm{800}}{{q}^{\mathrm{2}} } \\ $$$$\mathrm{21}−\mathrm{5cos2}\theta+\mathrm{16sin2}\theta=\frac{\mathrm{800}}{{q}^{\mathrm{2}} } \\ $$$$\mathrm{21}−\mathrm{5}\frac{\mathrm{1}−\mathrm{tg}^{\mathrm{2}} \theta}{\mathrm{1}+\mathrm{tg}^{\mathrm{2}} \theta}+\mathrm{16}\frac{\mathrm{2tg}\theta}{\mathrm{1}+\mathrm{tg}^{\mathrm{2}} \theta}=\frac{\mathrm{800}}{{q}^{\mathrm{2}} } \\ $$$$\left(\mathrm{26}−\frac{\mathrm{800}}{{q}^{\mathrm{2}} }\right)\mathrm{tg}^{\mathrm{2}} \theta+\mathrm{32tg}\theta+\left(\mathrm{16}−\frac{\mathrm{800}}{{q}^{\mathrm{2}} }\right)=\mathrm{0} \\ $$$$\mathrm{tg}\theta=\frac{−\mathrm{32}+\sqrt{\mathrm{1024}−\mathrm{4}\left(\mathrm{26}−\frac{\mathrm{800}}{{q}^{\mathrm{2}} }\right)\left(\mathrm{16}−\frac{\mathrm{800}}{{q}^{\mathrm{2}} }\right)}}{\mathrm{2}\left(\mathrm{26}−\frac{\mathrm{800}}{{q}^{\mathrm{2}} }\right)} \\ $$$${t}=\frac{\mathrm{400}}{{q}^{\mathrm{2}} } \\ $$$$\frac{−\mathrm{16}+\mathrm{2}\sqrt{−\mathrm{40}+\mathrm{21}{t}−{t}^{\mathrm{2}} }}{\mathrm{13}−{t}}=\frac{−\mathrm{10}−\mathrm{2}\sqrt{−\mathrm{40}+\mathrm{18}{t}−{t}^{\mathrm{2}} }}{\mathrm{5}−{t}} \\ $$$${t}=\frac{\mathrm{77}}{\mathrm{5}} \\ $$$${S}_{{ABCD}} ={q}^{\mathrm{2}} =\frac{\mathrm{400}}{{t}}=\frac{\mathrm{2000}}{\mathrm{77}} \\ $$$$ \\ $$
Commented by Lekhraj last updated on 26/May/25
Thank you
$$\mathrm{Thank}\:\mathrm{you}\: \\ $$
Answered by mr W last updated on 26/May/25
Commented by Lekhraj last updated on 26/May/25
Thank you
$$\mathrm{Thank}\:\mathrm{you} \\ $$
Commented by mr W last updated on 26/May/25
D(s sin θ, s sin θ+s cos θ) C(s cos θ+s sin θ, s cos θ) (((s sin θ)^2 )/a^2 )+(((s sin θ+s cos θ)^2 )/b^2 )=1 ⇒((sin^2 θ)/a^2 )+((1+sin 2θ)/b^2 )=(1/s^2 ) ...(i) (((s cos θ+s sin θ)^2 )/a^2 )+(((s cos θ)^2 )/b^2 )=1 ⇒((1+sin 2θ)/a^2 )+((cos^2 θ)/b^2 )=(1/s^2 ) ...(ii) ((1+sin 2θ)/a^2 )+((cos^2 θ)/b^2 )=((sin^2 θ)/a^2 )+((1+sin 2θ)/b^2 ) ((cos^2 θ+sin 2θ)/a^2 )=((sin^2 θ+sin 2θ)/b^2 ) (a^2 +b^2 ) cos 2θ=(a^2 −b^2 )(1+2 sin 2θ) with λ=((a^2 −b^2 )/(a^2 +b^2 )) cos 2θ=λ(1+2 sin 2θ) ((1−tan^2 θ)/(1+tan^2 θ))=λ(1+((4 tan θ)/(1+tan^2 θ))) (1+λ)tan^2 θ+4λ tan θ−(1−λ)=0 ⇒tan θ=((−2λ+(√(1+3λ^2 )))/(1+λ)) (1/s^2 )=((1+sin 2θ)/a^2 )+((cos^2 θ)/b^2 ) ((1+tan^2 θ)/s^2 )=(((1+tan θ)^2 )/a^2 )+(1/b^2 ) ⇒s^2 =((1+tan^2 θ)/((((1+tan θ)^2 )/a^2 )+(1/b^2 )))=area of square example: a^2 =80, b^2 =50 λ=((80−50)/(80+50))=(3/(13)) tan θ=((−(6/(13))+(√(1+3×(3^2 /(13^2 )))))/(1+(3/(13))))=(1/2) s^2 =((5/4)/((9/(4×80))+(1/(50))))=((2000)/(77)) ✓
$${D}\left({s}\:\mathrm{sin}\:\theta,\:{s}\:\mathrm{sin}\:\theta+{s}\:\mathrm{cos}\:\theta\right) \\ $$$${C}\left({s}\:\mathrm{cos}\:\theta+{s}\:\mathrm{sin}\:\theta,\:{s}\:\mathrm{cos}\:\theta\right) \\ $$$$\frac{\left({s}\:\mathrm{sin}\:\theta\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{\left({s}\:\mathrm{sin}\:\theta+{s}\:\mathrm{cos}\:\theta\right)^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\Rightarrow\frac{\mathrm{sin}^{\mathrm{2}} \:\theta}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}+\mathrm{sin}\:\mathrm{2}\theta}{{b}^{\mathrm{2}} }=\frac{\mathrm{1}}{{s}^{\mathrm{2}} }\:\:\:…\left({i}\right) \\ $$$$\frac{\left({s}\:\mathrm{cos}\:\theta+{s}\:\mathrm{sin}\:\theta\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{\left({s}\:\mathrm{cos}\:\theta\right)^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\Rightarrow\frac{\mathrm{1}+\mathrm{sin}\:\mathrm{2}\theta}{{a}^{\mathrm{2}} }+\frac{\mathrm{cos}^{\mathrm{2}} \:\theta}{{b}^{\mathrm{2}} }=\frac{\mathrm{1}}{{s}^{\mathrm{2}} }\:\:\:…\left({ii}\right) \\ $$$$\frac{\mathrm{1}+\mathrm{sin}\:\mathrm{2}\theta}{{a}^{\mathrm{2}} }+\frac{\mathrm{cos}^{\mathrm{2}} \:\theta}{{b}^{\mathrm{2}} }=\frac{\mathrm{sin}^{\mathrm{2}} \:\theta}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}+\mathrm{sin}\:\mathrm{2}\theta}{{b}^{\mathrm{2}} } \\ $$$$\frac{\mathrm{cos}^{\mathrm{2}} \:\theta+\mathrm{sin}\:\mathrm{2}\theta}{{a}^{\mathrm{2}} }=\frac{\mathrm{sin}^{\mathrm{2}} \:\theta+\mathrm{sin}\:\mathrm{2}\theta}{{b}^{\mathrm{2}} } \\ $$$$\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\:\mathrm{cos}\:\mathrm{2}\theta=\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)\left(\mathrm{1}+\mathrm{2}\:\mathrm{sin}\:\mathrm{2}\theta\right) \\ $$$${with}\:\lambda=\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$$\mathrm{cos}\:\mathrm{2}\theta=\lambda\left(\mathrm{1}+\mathrm{2}\:\mathrm{sin}\:\mathrm{2}\theta\right) \\ $$$$\frac{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:\theta}{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\theta}=\lambda\left(\mathrm{1}+\frac{\mathrm{4}\:\mathrm{tan}\:\theta}{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\theta}\right) \\ $$$$\left(\mathrm{1}+\lambda\right)\mathrm{tan}^{\mathrm{2}} \:\theta+\mathrm{4}\lambda\:\mathrm{tan}\:\theta−\left(\mathrm{1}−\lambda\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{tan}\:\theta=\frac{−\mathrm{2}\lambda+\sqrt{\mathrm{1}+\mathrm{3}\lambda^{\mathrm{2}} }}{\mathrm{1}+\lambda} \\ $$$$\frac{\mathrm{1}}{{s}^{\mathrm{2}} }=\frac{\mathrm{1}+\mathrm{sin}\:\mathrm{2}\theta}{{a}^{\mathrm{2}} }+\frac{\mathrm{cos}^{\mathrm{2}} \:\theta}{{b}^{\mathrm{2}} } \\ $$$$\frac{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\theta}{{s}^{\mathrm{2}} }=\frac{\left(\mathrm{1}+\mathrm{tan}\:\theta\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} } \\ $$$$\Rightarrow{s}^{\mathrm{2}} =\frac{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\theta}{\frac{\left(\mathrm{1}+\mathrm{tan}\:\theta\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }}={area}\:{of}\:{square} \\ $$$${example}: \\ $$$${a}^{\mathrm{2}} =\mathrm{80},\:{b}^{\mathrm{2}} =\mathrm{50} \\ $$$$\lambda=\frac{\mathrm{80}−\mathrm{50}}{\mathrm{80}+\mathrm{50}}=\frac{\mathrm{3}}{\mathrm{13}} \\ $$$$\mathrm{tan}\:\theta=\frac{−\frac{\mathrm{6}}{\mathrm{13}}+\sqrt{\mathrm{1}+\mathrm{3}×\frac{\mathrm{3}^{\mathrm{2}} }{\mathrm{13}^{\mathrm{2}} }}}{\mathrm{1}+\frac{\mathrm{3}}{\mathrm{13}}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${s}^{\mathrm{2}} =\frac{\frac{\mathrm{5}}{\mathrm{4}}}{\frac{\mathrm{9}}{\mathrm{4}×\mathrm{80}}+\frac{\mathrm{1}}{\mathrm{50}}}=\frac{\mathrm{2000}}{\mathrm{77}}\:\checkmark \\ $$
Commented by Lekhraj last updated on 26/May/25
Thank you
$$\mathrm{Thank}\:\mathrm{you}\: \\ $$
Answered by Spillover last updated on 27/May/25
Commented by Lekhraj last updated on 27/May/25
Thank you .
$$\mathrm{Thank}\:\mathrm{you}\:. \\ $$

Pin

Leave a Reply

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