The distance between the foci of the ellipse | vedclass

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(B) The given equation is $(3(x-3))^2 + 9y^2 = (\sqrt{2}x + y + 1)^2$.Dividing by $9$,we get $(x-3)^2 + y^2 = \frac{1}{9}(\sqrt{2}x + y + 1)^2$.This i

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$\frac{3\sqrt{2} + 1}{\sqrt{3}}$

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(B) The given equation is $(3(x-3))^2 + 9y^2 = (\sqrt{2}x + y + 1)^2$.Dividing by $9$,we get $(x-3)^2 + y^2 = \frac{1}{9}(\sqrt{2}x + y + 1)^2$.This is in the form $SP^2 = e^2 PM^2$,where $S(3, 0)$ is the focus and the directrix is $\sqrt{2}x + y + 1 = 0$.Here,$e^2 = \frac{1}{9} 	imes (	ext{normalization factor})^2$. The distance from $(x, y)$ to the line $\sqrt{2}x + y + 1 = 0$ is $\frac{|\sqrt{2}x + y + 1|}{\sqrt{(\sqrt{2})^2 + 1^2}} = \frac{|\sqrt{2}x + y + 1|}{\sqrt{3}}$.So,$(x-3)^2 + y^2 = \frac{1}{3} \left( \frac{\sqrt{2}x + y + 1}{\sqrt{3}} ight)^2$,which gives $e^2 = \frac{1}{3}$,so $e = \frac{1}{\sqrt{3}}$.The distance between the focus and the directrix is $\frac{a}{e} - ae = \frac{|\sqrt{2}(3) + 0 + 1|}{\sqrt{3}} = \frac{3\sqrt{2} + 1}{\sqrt{3}}$.Since $a(1/e - e) = \frac{3\sqrt{2} + 1}{\sqrt{3}}$,we have $a(\sqrt{3} - \frac{1}{\sqrt{3}}) = \frac{3\sqrt{2} + 1}{\sqrt{3}}$,so $a(\frac{2}{\sqrt{3}}) = \frac{3\sqrt{2} + 1}{\sqrt{3}}$,which gives $a = \frac{3\sqrt{2} + 1}{2}$.The distance between the foci is $2ae = 2 	imes \frac{3\sqrt{2} + 1}{2} 	imes \frac{1}{\sqrt{3}} = \frac{3\sqrt{2} + 1}{\sqrt{3}}$.

The distance between the foci of the ellipse $(3x - 9)^2 + 9y^2 = (\sqrt{2}x + y + 1)^2$ is:

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