If its eccentricity is the maximum value of the function,
Question:

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. (1) 145

  2. (2) 116

  3. (3) 126

  4. (4) 135


Correct Option: , 3

Solution:

The given ellipse is

$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1,(a>b)$

Length of latus rectum $=\frac{2 b^{2}}{a}$

$\Rightarrow \frac{2 b^{2}}{a}=10 \Rightarrow b^{2}=5 a$……(i)

Now $\phi(t)=\frac{5}{12}+t-t^{2}$

$\phi^{\prime}(t)=1-2 t=0 \Rightarrow t=\frac{1}{2}$

$\phi^{\prime \prime}(t)=-2<0 \Rightarrow$ maximum

$\Rightarrow \phi(t)_{\max }=\frac{5}{12}+\frac{1}{2}-\frac{1}{4}=\frac{8}{12}=\frac{2}{3}$

Since, $\phi(t)_{\max .}=$ eccentricity

$\Rightarrow e=\frac{2}{3}$

Now, $b^{2}=a^{2}\left(1-e^{2}\right)$

$5 a=a^{2}\left(1-\frac{4}{9}\right) \Rightarrow 5 a=\frac{5 a^{2}}{9} \Rightarrow a^{2}-9 a=0$

$\Rightarrow a=9 \Rightarrow a^{2}=81$ and $b^{2}=45$

$\therefore a^{2}+b^{2}=81+45=126$

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