If the focus of an ellipse is (-1, -1),the eq | vedclass

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(A) Let the focus be $S(-1, -1)$ and the directrix be $L: x + y + 1 = 0$. The eccentricity is $e = \frac{1}{\sqrt{2}}$.The distance $d$ between the fo

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$2$

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(A) Let the focus be $S(-1, -1)$ and the directrix be $L: x + y + 1 = 0$. The eccentricity is $e = \frac{1}{\sqrt{2}}$.The distance $d$ between the focus and the directrix is given by $d = \frac{|-1 - 1 + 1|}{\sqrt{1^2 + 1^2}} = \frac{1}{\sqrt{2}}$.For an ellipse,the distance between the focus and the directrix is $\frac{a}{e} - ae = d$.Substituting the values: $\frac{a}{1/\sqrt{2}} - a(\frac{1}{\sqrt{2}}) = \frac{1}{\sqrt{2}}$.$a\sqrt{2} - \frac{a}{\sqrt{2}} = \frac{1}{\sqrt{2}}$.Multiplying by $\sqrt{2}$: $2a - a = 1$,which gives $a = 1$.The length of the major axis is $2a = 2(1) = 2$.

If the focus of an ellipse is $(-1, -1)$,the equation of its directrix corresponding to this focus is $x + y + 1 = 0$,and its eccentricity is $e = \frac{1}{\sqrt{2}}$,then the length of its major axis is

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