If 3x + 4y = 12sqrt{2} is a tangent to the el | vedclass

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(B) The equation of the line is $3x + 4y = 12\sqrt{2}$,which can be rewritten as $y = -\frac{3}{4}x + 3\sqrt{2}$.Comparing this with $y = mx + c$,we h

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$2\sqrt{7}$

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(B) The equation of the line is $3x + 4y = 12\sqrt{2}$,which can be rewritten as $y = -\frac{3}{4}x + 3\sqrt{2}$.Comparing this with $y = mx + c$,we have $m = -\frac{3}{4}$ and $c = 3\sqrt{2}$.The condition for the line $y = mx + c$ to be a tangent to the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ is $c^{2} = a^{2}m^{2} + b^{2}$.Here,$b^{2} = 9$. Substituting the values: $(3\sqrt{2})^{2} = a^{2}(-\frac{3}{4})^{2} + 9$.$18 = a^{2}(\frac{9}{16}) + 9$.$9 = a^{2}(\frac{9}{16}) \Rightarrow a^{2} = 16$,so $a = 4$.The ellipse is $\frac{x^{2}}{16} + \frac{y^{2}}{9} = 1$. Since $a^{2} > b^{2}$,the eccentricity $e = \sqrt{1 - \frac{b^{2}}{a^{2}}} = \sqrt{1 - \frac{9}{16}} = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4}$.The distance between the foci is $2ae = 2 	imes 4 	imes \frac{\sqrt{7}}{4} = 2\sqrt{7}$.

If $3x + 4y = 12\sqrt{2}$ is a tangent to the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{9} = 1$ for some $a \in R$,then the distance between the foci of the ellipse is:

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