If ax^2 + by^2 = 15 is the equation of the el | vedclass

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(C) The given equation is $ax^2 + by^2 = 15$,which can be written as $\frac{x^2}{15/a} + \frac{y^2}{15/b} = 1$. Let $a'^2 = \frac{15}{a}$ and $b'^2 =

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$16$

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(C) The given equation is $ax^2 + by^2 = 15$,which can be written as $\frac{x^2}{15/a} + \frac{y^2}{15/b} = 1$. Let $a'^2 = \frac{15}{a}$ and $b'^2 = \frac{15}{b}$.For an ellipse,the distance between foci is $2a'e = 2 \Rightarrow a'e = 1$.The distance between directrices is $\frac{2a'}{e} = 5 \Rightarrow \frac{a'}{e} = \frac{5}{2}$.Multiplying these two equations: $(a'e) 	imes (a'/e) = 1 	imes \frac{5}{2} \Rightarrow a'^2 = \frac{5}{2}$.Since $a'^2 = \frac{15}{a}$,we have $\frac{15}{a} = \frac{5}{2} \Rightarrow a = 6$.Using $a'e = 1$,we get $e^2 = \frac{1}{a'^2} = \frac{2}{5}$.For an ellipse,$b'^2 = a'^2(1 - e^2) = \frac{5}{2}(1 - \frac{2}{5}) = \frac{5}{2} 	imes \frac{3}{5} = \frac{3}{2}$.Since $b'^2 = \frac{15}{b}$,we have $\frac{15}{b} = \frac{3}{2} \Rightarrow b = 10$.Therefore,$a + b = 6 + 10 = 16$.

If $ax^2 + by^2 = 15$ is the equation of the ellipse for which the distance between its foci is $2$ and the distance between its directrices is $5$,then $a + b =$

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