Question Number 40594 by ajfour last updated on 24/Jul/18
Commented by ajfour last updated on 24/Jul/18
$${Solution}\:{to}\:{Q}.\mathrm{40581} \\ $$
Answered by ajfour last updated on 24/Jul/18
![l=(√(b^2 −(a^2 /4))) =((√(4b^2 −a^2 ))/2) TM=rcot θ = (a/2)tan 30° = lcos 2θ ⇒ cos 2θ = (a/(2(√3)l)) cot θ =(√((1+cos 2θ)/(1−cos 2θ))) =(√((1+(a/(2(√3)l)))/(1−(a/(2(√3)l))))) r=(a/(2(√3)cot θ))=(a/(2(√3)))(√((1−(a/(2(√3)l)))/(1+(a/(2(√3)l))))) r=(a/(2(√3)))(√((2(√3)l−a)/(2(√3)l+a))) =(a/(2(√3)))(((2(√3)l−a)/(√(12l^2 −a^2 )))) r= (a/(2(√3)))((((√3)(√(4b^2 −a^2 ))−a)/(√(12b^2 −3a^2 −a^2 )))) ⇒ r=(a/(4(√3)))[(((√3)(√(4b^2 −a^2 ))−a)/(√(3b^2 −a^2 )))] same as r=(a/(4(√3)))[((3(4b^2 −a^2 )−a^2 )/(√(3b^2 −a^2 )))]((1/((√3)(√(4b^2 −a^2 ))+a))) =((a(√(3b^2 −a^2 )))/(3(√(4b^2 −a^2 ))+a(√3))) .](Q40597.png)
$${l}=\sqrt{{b}^{\mathrm{2}} −\frac{{a}^{\mathrm{2}} }{\mathrm{4}}}\:=\frac{\sqrt{\mathrm{4}{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }}{\mathrm{2}} \\ $$$${TM}={r}\mathrm{cot}\:\theta\:=\:\frac{{a}}{\mathrm{2}}\mathrm{tan}\:\mathrm{30}°\:=\:{l}\mathrm{cos}\:\mathrm{2}\theta \\ $$$$\Rightarrow\:\:\mathrm{cos}\:\mathrm{2}\theta\:=\:\frac{{a}}{\mathrm{2}\sqrt{\mathrm{3}}{l}} \\ $$$$\mathrm{cot}\:\theta\:=\sqrt{\frac{\mathrm{1}+\mathrm{cos}\:\mathrm{2}\theta}{\mathrm{1}−\mathrm{cos}\:\mathrm{2}\theta}}\:=\sqrt{\frac{\mathrm{1}+\frac{{a}}{\mathrm{2}\sqrt{\mathrm{3}}{l}}}{\mathrm{1}−\frac{{a}}{\mathrm{2}\sqrt{\mathrm{3}}{l}}}} \\ $$$${r}=\frac{{a}}{\mathrm{2}\sqrt{\mathrm{3}}\mathrm{cot}\:\theta}=\frac{{a}}{\mathrm{2}\sqrt{\mathrm{3}}}\sqrt{\frac{\mathrm{1}−\frac{{a}}{\mathrm{2}\sqrt{\mathrm{3}}{l}}}{\mathrm{1}+\frac{{a}}{\mathrm{2}\sqrt{\mathrm{3}}{l}}}} \\ $$$$\boldsymbol{{r}}=\frac{{a}}{\mathrm{2}\sqrt{\mathrm{3}}}\sqrt{\frac{\mathrm{2}\sqrt{\mathrm{3}}{l}−{a}}{\mathrm{2}\sqrt{\mathrm{3}}{l}+{a}}}\:=\frac{{a}}{\mathrm{2}\sqrt{\mathrm{3}}}\left(\frac{\mathrm{2}\sqrt{\mathrm{3}}{l}−{a}}{\sqrt{\mathrm{12}{l}^{\mathrm{2}} −{a}^{\mathrm{2}} }}\right) \\ $$$$\:{r}=\:\frac{{a}}{\mathrm{2}\sqrt{\mathrm{3}}}\left(\frac{\sqrt{\mathrm{3}}\sqrt{\mathrm{4}{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }−{a}}{\sqrt{\mathrm{12}{b}^{\mathrm{2}} −\mathrm{3}{a}^{\mathrm{2}} −{a}^{\mathrm{2}} }}\right) \\ $$$$\Rightarrow\:\:{r}=\frac{{a}}{\mathrm{4}\sqrt{\mathrm{3}}}\left[\frac{\sqrt{\mathrm{3}}\sqrt{\mathrm{4}{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }−{a}}{\sqrt{\mathrm{3}{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }}\right]\: \\ $$$${same}\:{as} \\ $$$$\:{r}=\frac{{a}}{\mathrm{4}\sqrt{\mathrm{3}}}\left[\frac{\mathrm{3}\left(\mathrm{4}{b}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)−{a}^{\mathrm{2}} }{\sqrt{\mathrm{3}{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }}\right]\left(\frac{\mathrm{1}}{\sqrt{\mathrm{3}}\sqrt{\mathrm{4}{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }+{a}}\right) \\ $$$$\:\:=\frac{{a}\sqrt{\mathrm{3}{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }}{\mathrm{3}\sqrt{\mathrm{4}{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }+{a}\sqrt{\mathrm{3}}}\:. \\ $$
Commented by MrW3 last updated on 24/Jul/18
![you′r answer is correct sir. ⇒ r=(a/(4(√3)))[(((√3)(√(4b^2 −a^2 ))−a)/(√(3b^2 −a^2 )))]×(((√3)(√(4b^2 −a^2 ))+a)/((√3)(√(4b^2 −a^2 ))+a)) ⇒ r=(a/(√3))[((3b^2 −a^2 )/(√(3b^2 −a^2 )))]×(1/((√3)(√(4b^2 −a^2 ))+a)) ⇒ r=((a(√(3b^2 −a^2 )))/((√3)a+3(√(4b^2 −a^2 )))) this is the same as my answer.](Q40608.png)
$${you}'{r}\:{answer}\:{is}\:{correct}\:{sir}. \\ $$$$\Rightarrow\:\:{r}=\frac{{a}}{\mathrm{4}\sqrt{\mathrm{3}}}\left[\frac{\sqrt{\mathrm{3}}\sqrt{\mathrm{4}{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }−{a}}{\sqrt{\mathrm{3}{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }}\right]×\frac{\sqrt{\mathrm{3}}\sqrt{\mathrm{4}{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }+{a}}{\sqrt{\mathrm{3}}\sqrt{\mathrm{4}{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }+{a}} \\ $$$$\Rightarrow\:\:{r}=\frac{{a}}{\sqrt{\mathrm{3}}}\left[\frac{\mathrm{3}{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\sqrt{\mathrm{3}{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }}\right]×\frac{\mathrm{1}}{\sqrt{\mathrm{3}}\sqrt{\mathrm{4}{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }+{a}} \\ $$$$\Rightarrow\:\:{r}=\frac{{a}\sqrt{\mathrm{3}{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }}{\sqrt{\mathrm{3}}{a}+\mathrm{3}\sqrt{\mathrm{4}{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }} \\ $$$${this}\:{is}\:{the}\:{same}\:{as}\:{my}\:{answer}. \\ $$
Commented by ajfour last updated on 24/Jul/18
$${i}'{ve}\:{even}\:{matched}\:{it}\:{myself},\:{Sir}. \\ $$$${thanks}\:{for}\:{the}\:{care}. \\ $$