Locate the complex numbers $z = x + i y$ such that
$| z - 1 | + | z + 1 | \leq 4$
Show that the point $z$ satisfying the above equation represents the interior and boundary ellipse $\frac{x^{2}}{4} + \frac{y^{2}}{3} = 1$ .

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Solution

$\sqrt{{(x - 1)}^{2} + y^{2}} + \sqrt{{(x + 1)}^{2} + y^{2}} \leq 4$
Consider $L + M = 4$
$L^{2} - M^{2} = - 4 x ∴ L - M = - x$
$∴ 2 L = 4 - x$ or $4 L^{2} = (4 - x)^{2}$
or $4 [x^{2} + y^{2} - 2 x + 1] = 16 + x^{2} - 8 x$
$3 x^{2} + 4 y^{2} = 12$ .
Hence the given equation is
$\frac{x^{2}}{4} + \frac{y^{2}}{3} - 1 \leq 0$ or $S^{'} \leq 0$ ...... $(1)$
$\frac{x^{2}}{4} + \frac{y^{2}}{3} - 1$ .

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