Consider an elipse having its foci at $A (z_{1}) and B (z_{2})$ in the argand plane. If the eccentricity of the ellipse is $e$ and it is known that origin is an interior point of the ellipse, then $e$ lies in the interval

(0, \frac{| z_{1} - z_{2} |}{| z_{1} | + | z_{2} |})

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(0, \frac{| z_{1} - z_{2} |}{| z_{1}^{2} | + | z_{2} |^{2}})

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(0, \frac{| z_{1} |}{| z_{2} |})

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(0, \frac{| z_{2} |}{| z_{1} |})

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Solution

The correct option is A $(0, \frac{| z_{1} - z_{2} |}{| z_{1} | + | z_{2} |})$
Let $P (z)$ be any point on the ellipse. Then equation of the ellipse is
$| z - z_{1} | + | z - z_{2} | = \frac{| z_{1} - z_{2} |}{e} \dots (1)$
If we replace
$z$ by $z_{1}$ or $z_{2}$ . L.H.S. of equation $(1)$ becomes $| z_{1} - z_{2} |$ . Thus for any interior point of the ellipse, we have
$| z - z_{1} | + | z - z_{2} | < \frac{| z_{1} - z_{2} |}{e}$
It is given that origin is an interior point of the ellipse
$| 0 - z_{1} | + | 0 - z_{2} | < \frac{| z_{1} - z_{2} |}{e}$
$\Rightarrow 0 < e < \frac{| z_{1} - z_{2} |}{| z_{1} | + | z_{2} |}$

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