JEE MAIN - Mathematics (2023 - 6th April Evening Shift - No. 10)
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 $$25\left(\beta^{2} x^{2}+\alpha^{2} y^{2}\right)=\alpha^{2} \beta^{2}$$ is :
$$\frac{\sqrt{129}}{12}$$
$$\frac{3 \sqrt{15}}{12}$$
$$\frac{\sqrt{119}}{12}$$
$$\frac{\sqrt{117}}{12}$$
Explanation
Let $E$ be the person speak, English
$\therefore n(E)=75$
and $H$ be the person speak Hindi
$\therefore n(H)=40$
Let number of persons who speak both English and Hindi are $t$.
_6th_April_Evening_Shift_en_10_1.png)
$\begin{array}{rlrl} &\therefore \alpha+t+\beta =100 ........(i) \\\\ & \alpha+t =75........(i) \\\\ &\text { and }\beta+t =40 ........(iii)\end{array}$
From Equations (i) and (ii), $\beta=25$
From Equations (i) and (iii), $\alpha=60$ and from Eq. (i), $t=15$
We have, equation of ellipse
$$ \begin{aligned} 25\left(\beta^2 x^2+\alpha^2 y^2\right) & =\alpha^2 \beta^2 \\\\ \Rightarrow 25\left(\frac{x^2}{\alpha^2}+\frac{y^2}{\beta^2}\right) & =1 \end{aligned} $$
$\begin{aligned} & \Rightarrow 25\left(\frac{x^2}{3600}+\frac{y^2}{625}\right)=1 \\\\ & \Rightarrow \frac{x^2}{144}+\frac{y^2}{25}=1 \\\\ & \therefore \text { Eccentricity, } e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{25}{144}}=\frac{\sqrt{119}}{12}\end{aligned}$
$\therefore n(E)=75$
and $H$ be the person speak Hindi
$\therefore n(H)=40$
Let number of persons who speak both English and Hindi are $t$.
$\begin{array}{rlrl} &\therefore \alpha+t+\beta =100 ........(i) \\\\ & \alpha+t =75........(i) \\\\ &\text { and }\beta+t =40 ........(iii)\end{array}$
From Equations (i) and (ii), $\beta=25$
From Equations (i) and (iii), $\alpha=60$ and from Eq. (i), $t=15$
We have, equation of ellipse
$$ \begin{aligned} 25\left(\beta^2 x^2+\alpha^2 y^2\right) & =\alpha^2 \beta^2 \\\\ \Rightarrow 25\left(\frac{x^2}{\alpha^2}+\frac{y^2}{\beta^2}\right) & =1 \end{aligned} $$
$\begin{aligned} & \Rightarrow 25\left(\frac{x^2}{3600}+\frac{y^2}{625}\right)=1 \\\\ & \Rightarrow \frac{x^2}{144}+\frac{y^2}{25}=1 \\\\ & \therefore \text { Eccentricity, } e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{25}{144}}=\frac{\sqrt{119}}{12}\end{aligned}$
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