The length of the latus rectum of the ellipse | vedclass

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(A) Given the distance from the centre to the focus is $ae = \sqrt{5}$,so $a^2e^2 = 5$. Since $e^2 = 1 - \frac{b^2}{a^2}$,we have $a^2(1 - \frac{b^2}{

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$7$

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(A) Given the distance from the centre to the focus is $ae = \sqrt{5}$,so $a^2e^2 = 5$. Since $e^2 = 1 - \frac{b^2}{a^2}$,we have $a^2(1 - \frac{b^2}{a^2}) = 5$,which implies $a^2 - b^2 = 5$,or $b^2 = a^2 - 5$. The length of the latus rectum is $\frac{2b^2}{a} = \frac{8}{3}$. Substituting $b^2 = a^2 - 5$,we get $\frac{2(a^2 - 5)}{a} = \frac{8}{3}$. $6(a^2 - 5) = 8a \Rightarrow 3a^2 - 4a - 15 = 0$. Solving the quadratic equation: $3a^2 - 9a + 5a - 15 = 0 \Rightarrow 3a(a - 3) + 5(a - 3) = 0$. Since $a > 0$,we have $a = 3$. Then $b^2 = 3^2 - 5 = 4$,so $b = 2$. Finally,$\sqrt{a^2 + 6ab + b^2} = \sqrt{3^2 + 6(3)(2) + 2^2} = \sqrt{9 + 36 + 4} = \sqrt{49} = 7$.

The length of the latus rectum of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ $(a > b)$ is $\frac{8}{3}$. If the distance from the centre of the ellipse to its focus is $\sqrt{5}$,then $\sqrt{a^2 + 6ab + b^2} =$

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