Solve this following


Let $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b)$ be a given ellipse, length of whose latus rectum is 10 . If its eccentricity is the maximum value of the

function, $\phi(t)=\frac{5}{12}+t-t^{2}$, then $a^{2}+b^{2}$ is equal to :


  1. 126

  2. 135

  3. 145

  4. 116

Correct Option: 1


$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b) ; \frac{2 b^{2}}{a}=10 \Rightarrow b^{2}=5 a \ldots(i)$

Now, $\phi(\mathrm{t})=\frac{5}{12}+\mathrm{t}-\mathrm{t}^{2}=\frac{8}{12}-\left(\mathrm{t}-\frac{1}{2}\right)^{2}$

$\phi(t)_{\max }=\frac{8}{12}=\frac{2}{3}=e \Rightarrow e^{2}=1-\frac{b^{2}}{a^{2}}=\frac{4}{9} \quad \ldots$ (ii)

$\Rightarrow a^{2}=81 \quad($ from (i) & (ii))

So, $a^{2}+b^{2}=81+45=126$


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