Find the equation of the ellipse whose latus | vedclass

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(A) Given that the length of the latus rectum is $\frac{2b^{2}}{a} = 10$ and the length of the minor axis $2b$ is equal to the distance between the fo

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$x^{2} + 2y^{2} = 100$

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(A) Given that the length of the latus rectum is $\frac{2b^{2}}{a} = 10$ and the length of the minor axis $2b$ is equal to the distance between the foci $2ae$.So,$2b = 2ae \Rightarrow b = ae$.Since $b^{2} = a^{2}(1 - e^{2})$,we have $(ae)^{2} = a^{2}(1 - e^{2})$,which implies $e^{2} = 1 - e^{2}$,so $2e^{2} = 1$,or $e = \frac{1}{\sqrt{2}}$.From $b = ae$,we have $b^{2} = a^{2}e^{2} = a^{2}(\frac{1}{2}) = \frac{a^{2}}{2}$.Substituting this into the latus rectum formula: $\frac{2(a^{2}/2)}{a} = 10 \Rightarrow a = 10$.Then $b^{2} = \frac{100}{2} = 50$.The equation of the ellipse is $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$,which is $\frac{x^{2}}{100} + \frac{y^{2}}{50} = 1$.Multiplying by $100$,we get $x^{2} + 2y^{2} = 100$.

Find the equation of the ellipse whose latus rectum is $10$ and the length of the minor axis is equal to the distance between the foci.

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