(182)

In right triangle OFP, we have $\tan 30°=\frac{PF}{OF}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{b^2}{a^{2}e} \Rightarrow e=\frac{\sqrt{3}{b}^2}{a^2}$

$e=\frac{\sqrt{3}{b}^2}{9}$          (Given $a^2=9$)

Squaring both sides, we get

$e^2=\frac{3b^4}{81}$          ... (i)

and  $e^2=1+\frac{b^2}{a^2} \Rightarrow e^2=1+\frac{b^2}{9}$          ... (ii)

Comparing equation (i) and (ii), we get

$\frac{3b^4}{81}=1+\frac{b^2}{9} \Rightarrow b^4=27+3b^2$

$\Rightarrow b^4–3b^2–27=0 \Rightarrow b^2=\frac{3}{2}(1+\sqrt{13})$

According to given condition, $b^2=\frac{l}{m}(1+\sqrt{n})$

On comparing, we get $l$ = 3, m = 2 and n = 13

$\Rightarrow l^2+m^2+n^2={\left(3\right)}^2+{\left(2\right)}^2+{\left(13\right)}^2=9+4+160=182$.

(51)

Here ae = 5 and $\frac{2b^2}{a}=\sqrt{50}$

     $\frac{b^2}{a}=\frac{5\sqrt{2}}{2}$

     $a^2{e}^2=a^2–b^2 \Rightarrow e^2=1–\frac{b^2}{a^2}$

     $25=a^2–\frac{5\sqrt{2}}{2}a \Rightarrow 2a^2–5\sqrt{2}a–50=0$

$\Rightarrow [a–5\sqrt{2}][2a+5\sqrt{2}]=0 \Rightarrow a=5\sqrt{2} \mathrm{or} a=\frac{–5\sqrt{2}}{2}$

$\Rightarrow e=\frac{5}{a}=\frac{1}{\sqrt{2}}$

Also, $b^2=5\sqrt{2}\times \frac{5\sqrt{2}}{2}=25 \Rightarrow b=5$

Now, $e=\sqrt{1+\frac{a^2{b}^2}{b^2}}=\sqrt{1+a^2} \Rightarrow e^2=1+a^2=51$.