Locus of foot of perpendicular drawn from the centre $(0, 0)$ to any tangent on the ellipse ${kx}^{2} + 3 y^{2} = 2$ satisfies the equation $3 {kx}^{2} + 4 y^{2} - 6 {(x^{2} + y^{2})}^{2} = 0$ then the value of $\frac{1}{e^{2}}$ (where e is eccentricity of the ellipse) is ……..

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Detailed Solution

$\begin{array}{l} \frac{x \cos ϕ}{a} + \frac{y \sin ϕ}{b} = 1 \\ \Rightarrow \frac{α \cos ϕ}{a} + \frac{β \sin ϕ}{b} = 1 . . . . (1) \\ & \frac{β}{α} (\frac{- bcos ϕ}{a \sin α}) = - 1 \dots . (2) \end{array}$
Solving 1&2 $(α^{2} + β^{2}) = a^{2} α^{2} + b^{2} β^{2} \Rightarrow$ equation of the required locus is ${(x^{2} + y^{2})}^{2} = a^{2} x^{2} + b^{2} y^{2}$
$\Rightarrow {(x^{2} + y^{2})}^{2} = \frac{x^{2}}{\frac{2}{K}} + \frac{y^{2}}{\frac{3}{K}}$
$\Rightarrow \frac{2}{K} = \frac{K}{2} \Rightarrow K = 2 \Rightarrow e = \sqrt{\frac{1 - \frac{2}{3}}{1}} = \frac{1}{\sqrt{3}}$