Let e1 and e2 be the eccentricities of the ellipse, x²/25 + y²/b² = 1(b

Q: Wait, why is the distance between the foci $2ae$ for both the ellipse and the hyperbola?

That is a fantastic observation! By definition, the foci of an ellipse are at $(\pm ae, 0)$, so the distance is $ae - (-ae) = 2ae$. Similarly, for a hyperbola in standard form, the foci are also at $(\pm ae, 0)$, making the distance $2ae$. The formula looks identical, but remember that the value of $e$ is calculated differently for each curve!

Q: Can I just assume $b^2$ is the same for both curves?

Yes, you absolutely can! The problem explicitly uses the same variable '$b^2$' in both equations. This is the 'bridge' that connects the two conic sections. Without this shared parameter, we would have two equations and too many unknowns to solve.

Q: Why did we divide by $b^2$ in the middle of the calculation? Is that safe?

Great catch! Dividing by a variable is only safe if we are certain it is not zero. In our case, if $b^2 = 0$, the ellipse would degenerate into a line segment, which contradicts the standard definition of an ellipse. Since $b^2 > 0$, we are mathematically justified in dividing by it to simplify our equation.

Q: Is there a physical meaning to $e_1 e_2 = 1$?

While it is primarily a mathematical constraint here, it implies a beautiful symmetry. It suggests that the 'flatness' of the ellipse is perfectly balanced by the 'spread' of the hyperbola. In coordinate geometry, such conditions often lead to elegant cancellations, which is exactly what we saw when the $1$s cancelled out!

Q: I got $b^2 = 9$. How do I know if this is correct?

Trust your algebra! You can verify it by plugging $b^2 = 9$ back into the original eccentricity formulas. If $e_1 = 4/5$ and $e_2 = 5/4$, their product is indeed $1$. Always perform this 'sanity check'—it is the hallmark of a top-tier JEE aspirant.

Question

Let $e_1$ and $e_2$ be the eccentricities of the ellipse, $\frac{x^2}{25} + \frac{y^2}{b^2} = 1(b < 5)$ and the hyperbola, $\frac{x^2}{16} - \frac{y^2}{b^2} = 1$ respectively satisfying $e_1e_2 = 1$. If $\alpha$ and $\beta$ are the distances between the foci of the ellipse and the foci of the hyperbola respectively, then the ordered pair $(\alpha, \beta)$ is equal to :

Options:

(A) (8,10)

(B) $(\frac{24}{5}, 10)$

(C) $(\frac{20}{3}, 12)$

(D) (8,12)

Sigma Percentile · Accepted Answer

### The Dance of the Conics: A Journey into Symmetry

Welcome, fellow traveler of the mathematical landscape. Today, we are not just solving a problem; we are exploring the elegant, hidden symmetry between two of the most fundamental shapes in the universe: the ellipse and the hyperbola.

Imagine you are standing in a coordinate plane. On one side, you have an ellipse, a closed, graceful loop. On the other, a hyperbola, a bold, sweeping curve that stretches toward infinity.

At first glance, they seem like opposites. But today, we will see how they are bound together by a single, beautiful condition: $e_1 e_2 = 1$.

### Phase 1: Defining the Players

Let us look at our equations. We have the ellipse:

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

And the hyperbola:

$$\frac{x^2}{16} - \frac{y^2}{b^2} = 1$$

Notice the shared parameter $b^2$. This is our anchor.

The eccentricity $e$ is the soul of a conic section; it tells us how much the curve deviates from a perfect circle. For our ellipse, the eccentricity is defined as:

$$e_1 = \sqrt{1 - \frac{b^2}{25}}$$

For our hyperbola, it is:

$$e_2 = \sqrt{1 + \frac{b^2}{16}}$$

### Phase 2: The Algebraic Bridge

We are given the condition $e_1 e_2 = 1$. When we multiply these two radicals, we get:

$$\sqrt{1 - \frac{b^2}{25}} \cdot \sqrt{1 + \frac{b^2}{16}} = 1$$

Now, I know that seeing square roots can be intimidating, but remember: the most powerful tool in your arsenal is the ability to simplify. Let us square both sides to liberate the terms from their radical prisons:

$$\left(1 - \frac{b^2}{25}ight) \left(1 + \frac{b^2}{16}ight) = 1$$

Expanding this, we get:

$$1 + \frac{b^2}{16} - \frac{b^2}{25} - \frac{b^4}{400} = 1$$

Look at that! The $1$ on both sides cancels out, leaving us with a beautiful, clean expression:

$$\frac{b^2}{16} - \frac{b^2}{25} - \frac{b^4}{400} = 0$$

By dividing by $b^2$ (since $b^2 
eq 0$), we find:

$$\frac{1}{16} - \frac{1}{25} = \frac{b^2}{400}$$

Taking the common denominator, we arrive at:

$$\frac{25 - 16}{400} = \frac{b^2}{400}$$

This simplifies instantly to **$b^2 = 9$**. The fog clears, and the path forward is illuminated.

### Phase 3: The Final Reveal

With $b^2 = 9$ in our hands, the rest of the problem unfolds like a well-choreographed dance. We calculate the eccentricities:

$$e_1 = \sqrt{1 - \frac{9}{25}} = \frac{4}{5}$$

$$e_2 = \sqrt{1 + \frac{9}{16}} = \frac{5}{4}$$

Finally, we calculate the distances between the foci. For the ellipse, the distance $\alpha$ is:

$$\alpha = 2 a_e e_1 = 2 \cdot 5 \cdot \frac{4}{5} = 8$$

For the hyperbola, the distance $\beta$ is:

$$\beta = 2 a_h e_2 = 2 \cdot 4 \cdot \frac{5}{4} = 10$$

Thus, the ordered pair $(\alpha, \beta)$ is **$(8, 10)$**.

Take a moment to appreciate this. We started with two distinct curves and a single constraint, and through logical deduction, we uncovered their exact dimensions. This is the beauty of mathematics—it turns chaos into order.

Select Answer:

Visualized Solution

Visualizing the Conics

Eccentricity of the Ellipse ( $e_{1}$ )

Eccentricity of the Hyperbola ( $e_{2}$ )

Applying the Condition $e_{1} e_{2} = 1$

Squaring Both Sides

Expanding the Equation

Solving for $b^{2}$

Calculating the Value of $b^{2}$

Calculating Exact Eccentricities

Distance Between Foci of Ellipse ( $α$ )

Distance Between Foci of Hyperbola ( $β$ )

Final Ordered Pair $(α, β)$

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

The Dance of the Conics

Phase 1

Phase 2

Phase 3

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Select Answer:

Visualized Solution

Visualizing the Conics

Eccentricity of the Ellipse ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } e1​)

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