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In a group of 100 persons 75 speak English and 40 speak Hindi. Each person speaks at least one of the two
languages. If the number of persons, who speak only English is $\alpha$ and the number of persons who speak only
Hindi is $\beta$ , then the eccentricity of the ellipse $\mathrm{25\left (\beta ^{2}x^{2}+\alpha ^{2}+y^{2} \right )=\alpha ^{2}\beta ^{2}}$ is :
Option: 1
$\frac{\sqrt{129}}{12}$

Option: 2
$\frac{\sqrt{117}}{12}$

Option: 3
$\frac{\sqrt{119}}{12}$

Option: 4
$\frac{3\sqrt{15}}{12}$

Answers (1)

best_answer

$\mathrm{n}(\mathrm{A} \cap \mathrm{B})=\mathrm{n}(\mathrm{A})+\mathrm{n}(\mathrm{B})-\mathrm{n}(\mathrm{A} \cap \mathrm{B}) \\$
$\mathrm{n}(\mathrm{A} \cap \mathrm{B})=75+40-100$
$\mathrm{n}(\mathrm{A} \cap \mathrm{B})=15 \alpha=60$
$\mathrm{\text { Only } \mathrm{E} \rightarrow 60 \beta=25 }$
$\mathrm{\text { Only } \mathrm{H} \rightarrow 25 }$
$\quad \text { Both }=15$
$\frac{25 \mathrm{x}^2}{\alpha^2}+\frac{25 \mathrm{y}^2}{\beta^2}=1$
$\frac{25 \mathrm{x}^2}{(60)^2}+\frac{\left(25 \mathrm{y}^2\right)}{(25)^2}=1$
$\mathrm{e}^2=1-\left[\frac{25 \times 25}{(60)^2}\right]$

$\mathrm{e^{2}=\frac{\left ( 60 \right )^{2}-\left ( 25 \right )^{2}}{\left ( 60 \right )^{2}}}$

$\mathrm{e^{2}=\frac{\left ( 60 -25\right )-\left ( 60+25 \right )}{ 60 \times 60}}$

$\mathrm{e^{2}=\frac{\left ( 35 \right )\left ( 85 \right )}{60\times 60}}=\frac{119}{144}$

$\mathrm{e=\frac{\sqrt{119}}{12}}$

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