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If $P$ lies in the first quadrant on the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ (where $a > b$ ), and tangent & normal drawn at $P$ meets major axis at the points $T$ & $N$ respectively, then the value of $\frac{{\left( {\left| {{F_2}N} \right| + \left| {{F_1}N} \right|} \right)\left( {\left| {{F_2}T} \right| - \left| {{F_1}T} \right|} \right)}}{{\left( {\left| {{F_2}N} \right| - \left| {{F_1}N} \right|} \right)\left( {\left| {{F_2}T} \right| + \left| {{F_1}T} \right|} \right)}}$ is equal to (where $F_1$ & $F_2$ are the foci $(ae, 0)$ & $(-ae, 0)$ respectively)
$1$
$2a$
$2b$
$\frac{a}{e}$
Solution

As in and $PT$ are the angle bisector of $\angle \mathrm{F}_{1}\,\, \&\,\, \mathrm{F}_{2}$ (reflection property)
then $\frac{\left|\mathrm{F}_{2} \mathrm{P}\right|}{\left|\mathrm{F}_{1} \mathrm{P}\right|}=\frac{\left|\mathrm{F}_{2} \mathrm{N}\right|}{\left|\mathrm{F}_{1} \mathrm{N}\right|}=\frac{\left|\mathrm{F}_{2} \mathrm{T}\right|}{\left|\mathrm{F}_{1} \mathrm{T}\right|}$
Using componendo and dividendo
$\left(\frac{\left|\mathrm{F}_{2} \mathrm{N}\right|+\left|\mathrm{F}_{1} \mathrm{N}\right|}{\left|\mathrm{F}_{2} \mathrm{N}\right|-\left|\mathrm{F}_{1} \mathrm{N}\right|}\right)\left(\frac{\left|\mathrm{F}_{2} \mathrm{T}\right|-\left|\mathrm{F}_{1} \mathrm{T}\right|}{\left|\mathrm{F}_{2} \mathrm{T}\right|+\left|\mathrm{F}_{1} \mathrm{T}\right|}\right)=1$