Let frac{x^{2}}{a^{2}}+frac{y^{2}}{b^{2}}=1(a | vedclass

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(A) Given the ellipse equation $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ with $a > b$.The length of the latus rectum is given by $\frac{2b^{2}}{

Vedclass · Accepted Answer

$126$

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(A) Given the ellipse equation $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ with $a > b$.The length of the latus rectum is given by $\frac{2b^{2}}{a} = 10$,which implies $b^{2} = 5a$ ... $(i)$.Now,consider the function $\phi(t) = \frac{5}{12} + t - t^{2}$.To find the maximum value,we complete the square: $\phi(t) = -\left(t^{2} - t + \frac{1}{4}ight) + \frac{1}{4} + \frac{5}{12} = -\left(t - \frac{1}{2}ight)^{2} + \frac{8}{12} = -\left(t - \frac{1}{2}ight)^{2} + \frac{2}{3}$.The maximum value is $\phi(t)_{	ext{max}} = \frac{2}{3}$,so $e = \frac{2}{3}$.Since $e^{2} = 1 - \frac{b^{2}}{a^{2}}$,we have $\frac{4}{9} = 1 - \frac{b^{2}}{a^{2}}$,which implies $\frac{b^{2}}{a^{2}} = \frac{5}{9}$,so $b^{2} = \frac{5}{9}a^{2}$ ... $(ii)$.Equating $(i)$ and $(ii)$,$5a = \frac{5}{9}a^{2} \Rightarrow a = \frac{a^{2}}{9} \Rightarrow a = 9$.Then $a^{2} = 81$ and $b^{2} = 5(9) = 45$.Therefore,$a^{2} + b^{2} = 81 + 45 = 126$.

Let $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b)$ be a given ellipse whose latus rectum length is $10$. If its eccentricity $e$ is the maximum value of the function $\phi(t) = \frac{5}{12} + t - t^{2}$,then $a^{2} + b^{2}$ is equal to:

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