Let the latus rectum of the hyperbola [(x^2)/(9)-(y^2)/(b^2)=1] subtend an angle of [()/(3…

Question

Let the latus rectum of the hyperbola $$\frac{x^2}{9}-\frac{y^2}{b^2}=1$$ subtend an angle of $$\frac{\pi}{3}$$ at the centre of the hyperbola. If $$\mathrm{b}^2$$ is equal to $$\frac{l}{\mathrm{~m}}(1+\sqrt{\mathrm{n}})$$, where $l$ and ...

Answer: 182

Accepted Answer

LR subtends ${60}^{\circ }$ at centre

$\begin{gathered} \Rightarrow 	an {30}^{\circ }=\frac{{b}^{2}/a}{ae}=\frac{{b}^{2}}{{a}^{2}e}=\frac{1}{\sqrt{3}} \ \Rightarrow e=\frac{\sqrt{3}{b}^{2}}{9} \end{gathered}$

Also, ${e}^{2}=1+\frac{{b}^{2}}{9}\Rightarrow 1+\frac{{b}^{2}}{9}=\frac{3{b}^{4}}{81}$

$\begin{gathered} \Rightarrow {b}^{4}=3{b}^{2}+27 \ \Rightarrow {b}^{4}-3{b}^{2}-27=0 \ \Rightarrow {b}^{2}=\frac{3}{2}(1+\sqrt{13}) \ \Rightarrow l=3,m=2,n=13 \ \Rightarrow {l}^{2}+{m}^{2}+{n}^{2}=182 \end{gathered}$

Solutions · 1