Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

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

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

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

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

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

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

Posted by

vinayak

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

50 mL of 0.2 M ammonia solution is treated with 25 mL of 0.2 M HCl. If pK_b of ammonia solution is 4.75, the pH of the mixture will be :
Option: 1 3.75
Option: 2 4.75

Latest Question