In a group of 100 persons 75 speak English and 40 speak Hindi. Each person speaks at least one of th

Q: Why do we subtract the intersection to find $\alpha$?

Great question! The set $n(E)$ represents all English speakers, which includes those who speak *only* English and those who speak *both* English and Hindi. Since $\alpha$ is defined as the number of people who speak *only* English, we must remove the bilingual group from the total English-speaking population. Mathematically, $\alpha = n(E) - n(E \cap H)$.

Q: How do I know if $a^2$ is under $x^2$ or $y^2$?

Always look at the denominators in the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. By definition, $a^2$ is the larger value. If the larger denominator is under $x^2$, the ellipse is horizontal (stretched along the x-axis). If it is under $y^2$, it is vertical. In this problem, after simplification, you will find the larger denominator is under $x^2$.

Q: Is there a shortcut for the eccentricity?

While there is no 'shortcut' that bypasses the calculation, always remember the relationship $e = \sqrt{1 - \frac{b^2}{a^2}}$ for a horizontal ellipse. The most common mistake is swapping $a^2$ and $b^2$. Just remember: $a^2$ is the larger one, so the fraction $\frac{b^2}{a^2}$ is always less than 1, ensuring the value under the square root is positive!

Question

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 speaks only Hindi is $\beta$, then the eccentricity of the ellipse $25(\beta^2 x^2 + \alpha^2 y^2) = \alpha^2 \beta^2$ is

Options:

(A) $\frac{\sqrt{119}}{12}$

(B) $\frac{\sqrt{117}}{12}$

(C) $\frac{3\sqrt{15}}{12}$

(D) $\frac{\sqrt{129}}{12}$

Sigma Percentile · Accepted Answer

### The Beautiful Bridge Between Sets and Conics

My dear student, welcome to a problem that is truly a masterclass in the elegance of mathematics. Often, we treat topics like Set Theory and Coordinate Geometry as separate islands in our JEE preparation.

But today, we are going to build a bridge between them. We are going to take the discrete, logical world of Venn diagrams and use it to fuel the continuous, graceful curves of an ellipse. Let us embark on this journey together.

### Phase 1: The Logic of the Overlap

Imagine you are standing in a room with one hundred people. You are told that seventy-five of them speak English and forty speak Hindi.

If you simply add these numbers, you get one hundred and fifteen. But wait—there are only one hundred people in the room! Where did the extra fifteen come from?

They are the bilingual souls, counted twice because they belong to both groups. This is the essence of the **Principle of Inclusion-Exclusion**.

We define the union of our sets as $n(E \cup H) = 100$, the English speakers as $n(E) = 75$, and the Hindi speakers as $n(H) = 40$. The formula is:

$$n(E \cup H) = n(E) + n(H) - n(E \cap H)$$

Substituting our values, we get $100 = 75 + 40 - n(E \cap H)$, which reveals that $n(E \cap H) = 15$. These fifteen people are the bridge between our two sets.

### Phase 2: Defining the Parameters

Now, we need to isolate the specific groups. $\alpha$ represents those who speak *only* English. We take the total English speakers and strip away the bilinguals: $\alpha = 75 - 15 = 60$.

Similarly, $\beta$ represents those who speak *only* Hindi: $\beta = 40 - 15 = 25$. We have successfully distilled our set theory problem into two clean, solid numbers: $\alpha = 60$ and $\beta = 25$.

These are the building blocks for our geometric masterpiece.

### Phase 3: The Geometry of the Ellipse

Now, the problem shifts. We are given the equation $25(\beta^2 x^2 + \alpha^2 y^2) = \alpha^2 \beta^2$. It looks daunting, but do not let the variables intimidate you.

Let us substitute our values: $25(25^2 x^2 + 60^2 y^2) = 60^2 \cdot 25^2$. To find the eccentricity, we need this in the standard form:

$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$

We divide the entire equation by the constant on the right, $60^2 \cdot 25^2$. This yields:

$$\frac{25 \cdot 25^2 x^2}{60^2 \cdot 25^2} + \frac{25 \cdot 60^2 y^2}{60^2 \cdot 25^2} = 1$$

Watch how the terms cancel out with such satisfying precision:

$$\frac{25 x^2}{60^2} + \frac{25 y^2}{25^2} = 1$$

Simplifying the denominators, we get $\frac{x^2}{(60^2/25)} + \frac{y^2}{25} = 1$. Since $60/5 = 12$, the first denominator is $12^2 = 144$. Thus, our equation is:

$$\frac{x^2}{144} + \frac{y^2}{25} = 1$$

### Phase 4: The Final Flourish

We have arrived at the standard form. Comparing this to $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, we identify $a^2 = 144$ and $b^2 = 25$.

Since $a^2 > b^2$, we are dealing with a horizontal ellipse. The eccentricity $e$ is defined by the beautiful formula:

$$e = \sqrt{1 - \frac{b^2}{a^2}}$$

Substituting our values, we get $e = \sqrt{1 - \frac{25}{144}}$. A quick calculation gives us:

$$e = \sqrt{\frac{144 - 25}{144}} = \sqrt{\frac{119}{144}}$$

Finally, taking the square root of the denominator, we find the **final answer**:

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

You see? By staying calm and following the logical flow, we transformed a set theory puzzle into a coordinate geometry triumph. Keep this clarity of thought, and no JEE problem will ever be able to stand against you.

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Visualized Solution

$V i s u a l i z in g t h e S e t s$

$F in d in g t h e I n t er sec t i o n$

$C a l c u l a t in g n (E \cap H)$

$C a l c u l a t in g α (O n l y E n g l i s h)$

$C a l c u l a t in g β (O n l y H in d i)$

$T h e E l l i p se E q u a t i o n$

$S im pl i f y in g t o S t an d a r d F or m$

$I d e n t i f y in g a^{2} an d b^{2}$

$C a l c u l a t in g E cce n t r i c i t y e$

$F ina l C o n c l u s i o n$

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

The Beautiful Bridge Between Sets and Conics

Phase 1

Phase 2

Phase 3

Phase 4

Similar Questions

Let $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1, a > b$ be an ellipse, whose eccentricity is $\frac{1}{2}$ and the length of the latus rectum is $14$ . Then the square of the eccentricity of $\frac{x ^{2}}{a ^{2}} - \frac{m y ^{2}}{b ^{2}} = 1$ is :

Let the eccentricity of an ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1, a > b$ , be $\frac{1}{4}$ . If this ellipse passes through the point $(- 4 \frac{2}{5}, 3)$ , then $a^{2} + b^{2}$ is equal to :

Let the maximum area of the triangle that can be inscribed in the ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{4} = 1, a > 2$ , having one of its vertices at one end of the major axis of the ellipse and one of its sides parallel to the y-axis, be $63$ . Then the eccentricity of the ellipse is :

The ellipse $E_{1} : \frac{x ^{2}}{9} + \frac{y ^{2}}{4} = 1$ is inscribed in a rectangle $R$ whose sides are parallel to the coordinate axes. Another ellipse $E_{2}$ passing through the point $(0, 4)$ circumscribes the rectangle $R$ . The eccentricity of the ellipse $E_{2}$ is

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Sets

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Finding the Intersection

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating n(E∩H)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating α (Only English)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating β (Only Hindi)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Ellipse Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Simplifying to Standard Form

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Identifying a2 and b2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Eccentricity e

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Conclusion