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QJEEM23Ma7738: JEE Main 2023 Mathematics - Class 11 JEEM23Ma7738 Solution: In a group of 100 persons 75 speak English and 40 speak Hindi.…

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 α\alpha and the number of persons who speak only Hindi is β\beta , then the eccentricity of the ellipse 25(β2x2+α2y2)=α2β225\left(\beta^{2} x^{2}+\alpha^{2} y^{2}\right)=\alpha^{2} \beta^{2} is :

Multiple Choice Options

  • A 12912\frac{\sqrt{129}}{12}
  • B 31512\frac{3 \sqrt{15}}{12}
  • C 11912\frac{\sqrt{119}}{12}
  • D 11712\frac{\sqrt{117}}{12}

Step-by-Step Verified Solution

Let EE be the person speak, English

∴n(E)=75\therefore n(E)=75

and HH be the person speak Hindi

∴n(H)=40\therefore n(H)=40

Let number of persons who speak both English and Hindi are tt .

JEE Main 2023 (Online) 6th April Evening Shift Mathematics - Ellipse Question 17 English Explanation
amp;∴α+t+β=100........(i)amp;α+t=75........(i)amp; and β+t=40........(iii)\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), β=25\beta=25

From Equations (i) and (iii), α=60\alpha=60 and from Eq. (i), t=15t=15

We have, equation of ellipse

25(β2x2+α2y2)amp;=α2β2⇒25(x2α2+y2β2)amp;=1\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}

amp;⇒25(x23600+y2625)=1amp;⇒x2144+y225=1amp;∴ Eccentricity, e=1−b2a2=1−25144=11912\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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