Explanation

(b)
Given equation of ellipse is
$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1(a < b)$
It is a vertical ellipse with foci $(0,\ \pm be)$
Equation of any tangent line to the above ellipse is
$y = mx + \sqrt{a^{2}m^{2} + b^{2}}$
$\therefore$ Required product
$= \left| \frac{- be + \sqrt{a^{2}m^{2} + b^{2}}}{\sqrt{m^{2} + 1}} \right|\left| \frac{be + \sqrt{a^{2}m^{2} + b^{2}}}{\sqrt{m^{2} + 1}} \right|$
$= \left| \frac{a^{2}m^{2} + b^{2} - b^{2}e^{2}}{m^{2} + 1} \right|$
$= \left| \frac{a^{2}m^{2} + b^{2}\left( 1 - e^{2} \right)}{m^{2} + 1} \right|$
$= \left| \frac{a^{2}m^{2} + a^{2}}{m^{2} + 1} \right|$ $\lbrack\because a^{2} = b^{2}(1 - e^{2})\rbrack$
$= a^{2}$