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(\beta^2 x^2+\alpha^2 y^2)=\alpha^2 \beta^2 is :

Ellipse PYQ — JEE Main 2023 · 6 Apr · Shift 2 · Q33 with Solution

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Correct answer: C

Find $\alpha$ and $\beta$ using set theory.

Let:

$E$ = persons speaking English, $|E|=75$
$H$ = persons speaking Hindi, $|H|=40$
Total persons $=100$

Since each person speaks at least one language, $|E\cup H|=100$

Using $|E\cup H|=|E|+|H|-|E\cap H|,$ we get $100=75+40-|E\cap H|$ $|E\cap H|=15$

Therefore,

persons speaking only English: $\alpha=75-15=60$
persons speaking only Hindi: $\beta=40-15=25$

Substitute into the ellipse equation.

Given: $25\left(\beta^{2}x^{2}+\alpha^{2}y^{2}\right)=\alpha^{2}\beta^{2}$

Substitute $\alpha=60,\ \beta=25$ : $25\left(25^{2}x^{2}+60^{2}y^{2}\right)=60^{2}\cdot 25^{2}$

Divide both sides by $25$ : $25^{2}x^{2}+60^{2}y^{2}=60^{2}\cdot 25$

Now divide by $60^{2}\cdot 25$ : $\frac{x^{2}}{60^{2}/25}+\frac{y^{2}}{25}=1$

A cleaner way is to simplify directly: $\beta^{2}x^{2}+\alpha^{2}y^{2}=\frac{\alpha^{2}\beta^{2}}{25}$ $\frac{x^{2}}{\alpha^{2}/25}+\frac{y^{2}}{\beta^{2}/25}=1$

So,

\qquad b^{2}=\frac{\beta^{2}}{25}=\frac{25^{2}}{25}=25$$ Thus, $$a=12,\qquad b=5$$ 3. **Find the eccentricity** of the ellipse. For ellipse, $$e=\sqrt{1-\frac{b^{2}}{a^{2}}}$$ So, $$e=\sqrt{1-\frac{25}{144}}=\sqrt{\frac{119}{144}}=\frac{\sqrt{119}}{12}$$ 4. **Match with options.** $$\boxed{\frac{\sqrt{119}}{12}}$$ which is **Option C**.

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View written solutionFree

Correct answer: C

Find $\alpha$ and $\beta$ using set theory.

Let:

$E$ = persons speaking English, $|E|=75$
$H$ = persons speaking Hindi, $|H|=40$
Total persons $=100$

Since each person speaks at least one language, $|E\cup H|=100$

Using $|E\cup H|=|E|+|H|-|E\cap H|,$ we get $100=75+40-|E\cap H|$ $|E\cap H|=15$

Therefore,

persons speaking only English: $\alpha=75-15=60$
persons speaking only Hindi: $\beta=40-15=25$

Substitute into the ellipse equation.

Given: $25\left(\beta^{2}x^{2}+\alpha^{2}y^{2}\right)=\alpha^{2}\beta^{2}$

Substitute $\alpha=60,\ \beta=25$ : $25\left(25^{2}x^{2}+60^{2}y^{2}\right)=60^{2}\cdot 25^{2}$

Divide both sides by $25$ : $25^{2}x^{2}+60^{2}y^{2}=60^{2}\cdot 25$

Now divide by $60^{2}\cdot 25$ : $\frac{x^{2}}{60^{2}/25}+\frac{y^{2}}{25}=1$

A cleaner way is to simplify directly: $\beta^{2}x^{2}+\alpha^{2}y^{2}=\frac{\alpha^{2}\beta^{2}}{25}$ $\frac{x^{2}}{\alpha^{2}/25}+\frac{y^{2}}{\beta^{2}/25}=1$

So,

\qquad b^{2}=\frac{\beta^{2}}{25}=\frac{25^{2}}{25}=25$$ Thus, $$a=12,\qquad b=5$$ 3. **Find the eccentricity** of the ellipse. For ellipse, $$e=\sqrt{1-\frac{b^{2}}{a^{2}}}$$ So, $$e=\sqrt{1-\frac{25}{144}}=\sqrt{\frac{119}{144}}=\frac{\sqrt{119}}{12}$$ 4. **Match with options.** $$\boxed{\frac{\sqrt{119}}{12}}$$ which is **Option C**.

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