An ellipse E : x²/a² + y²/b² = 1 passes through the vertices of the hyperbola H : x²/49 - y²/64 = -1

Q: Wait, why is the hyperbola vertical? I thought the standard form was always $x^2/a^2 - y^2/b^2 = 1$.

That is a fantastic question! The standard form $x^2/a^2 - y^2/b^2 = 1$ describes a horizontal hyperbola. However, when you see a $-1$ on the right side, it is a signal to multiply the entire equation by $-1$. This transforms it into $y^2/b^2 - x^2/a^2 = 1$. In the coordinate plane, this flips the hyperbola to open along the y-axis. Always check the sign on the right side before you start calculating vertices!

Q: Why did we use $e^2 = 1 - a^2/b^2$ for the ellipse? I usually use $1 - b^2/a^2$.

You are absolutely right to be cautious! The formula $e^2 = 1 - b^2/a^2$ is for a horizontal ellipse where $a > b$. But here, because the ellipse passes through the vertices of a vertical hyperbola, its major axis must be the y-axis. For a vertical ellipse, the semi-major axis is $b$, and the semi-minor axis is $a$. Thus, the formula becomes $e^2 = 1 - (\text{semi-minor})^2 / (\text{semi-major})^2$, which is $1 - a^2/b^2$. Always identify which axis is the major one first!

Q: Is there a shortcut for the latus rectum? The calculation $113l$ looks intimidating.

The beauty of JEE Advanced problems is that they are designed to simplify if you keep your expressions in their 'raw' form. Notice how we kept $a^2$ as a fraction involving $113$ instead of calculating the decimal? When you multiply by $113$ at the very end, the denominator cancels out perfectly. Don't rush to calculate decimals; keep your algebra clean, and the numbers will often 'talk' to you and simplify themselves.

Question

An ellipse $E : \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ passes through the vertices of the hyperbola $H : \frac{x^2}{49} - \frac{y^2}{64} = -1$. Let the major and minor axes of the ellipse $E$ coincide with the transverse and conjugate axes of the hyperbola $H$. Let the product of the eccentricities of $E$ and $H$ be $1/2$. If $l$ is the length of the latus rectum of the ellipse $E$, then the value of $113l$ is equal to ______.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, my fellow traveler in the world of mathematics. Today, we are not just solving a problem; we are exploring the elegant, interconnected geometry of conic sections.

Often, students look at a problem involving an ellipse and a hyperbola and see two separate, disconnected entities. But in the eyes of a mathematician, these curves are part of a beautiful, unified family. Let us peel back the layers of this problem together.

### The Hyperbola's Hidden Orientation

We begin with the hyperbola $H : \frac{x^2}{49} - \frac{y^2}{64} = -1$. The moment you see this, I want you to pause. Do not rush to calculate. Look at the right-hand side; it is $-1$.

This is the 'trap' that separates the casual student from the master. A standard horizontal hyperbola has a $+1$ on the right. By multiplying the entire equation by $-1$, we reveal its true nature:

$$ \frac{y^2}{64} - \frac{x^2}{49} = 1 $$

This is a vertical hyperbola, opening its arms towards the positive and negative y-axis. By setting $x=0$, we find the vertices at $y^2 = 64$, which gives us $y = \pm 8$. These two points, $(0, 8)$ and $(0, -8)$, are the anchors of our entire problem.

### The Geometric Bridge

Now, consider the ellipse $E : \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. We are told it passes through the vertices of the hyperbola. Since the hyperbola's vertices lie on the y-axis, the ellipse must also pass through $(0, 8)$ and $(0, -8)$.

Substituting these coordinates into the ellipse equation, we get:

$$ \frac{0^2}{a^2} + \frac{8^2}{b^2} = 1 \Rightarrow b^2 = 64 $$

This is a profound realization. Because the ellipse is centered at the origin and passes through $(0, \pm 8)$, and its axes coincide with the hyperbola's, we know that the semi-major axis of our ellipse is $b=8$, lying along the y-axis. Our ellipse is vertical, just like our hyperbola.

### The Eccentricity Dance

Next, we enter the realm of eccentricity—the 'DNA' of a conic section. It defines how 'stretched' or 'squashed' a curve is. For our vertical hyperbola, the eccentricity $e_H$ is given by the formula:

$$ e_H = \sqrt{1 + \frac{a_H^2}{b_H^2}} $$

Substituting our values, $a_H^2 = 49$ and $b_H^2 = 64$, we find:

$$ e_H = \sqrt{1 + \frac{49}{64}} = \sqrt{\frac{113}{64}} = \frac{\sqrt{113}}{8} $$

Now, the problem gives us a beautiful condition: the product of the eccentricities is $1/2$. So, $e_E \cdot e_H = \frac{1}{2}$. Substituting our value for $e_H$, we get $e_E \cdot \frac{\sqrt{113}}{8} = \frac{1}{2}$.

Solving for $e_E$, we find $e_E = \frac{4}{\sqrt{113}}$. This is the key that unlocks the final door.

### The Final Calculation

We are in the home stretch. For a vertical ellipse, the relationship between eccentricity and the semi-axes is $e_E^2 = 1 - \frac{a^2}{b^2}$. We know $e_E^2 = \frac{16}{113}$ and $b^2 = 64$.

Substituting these, we have:

$$ \frac{16}{113} = 1 - \frac{a^2}{64} $$

Rearranging this, we get $\frac{a^2}{64} = 1 - \frac{16}{113} = \frac{97}{113}$. Thus, $a^2 = \frac{64 \cdot 97}{113}$.

The length of the latus rectum $l$ for a vertical ellipse is defined as $l = \frac{2a^2}{b}$. Substituting our values, we get:

$$ l = \frac{2}{8} \cdot \left( \frac{64 \cdot 97}{113} \right) = \frac{1}{4} \cdot \frac{64 \cdot 97}{113} = \frac{16 \cdot 97}{113} = \frac{1552}{113} $$

Finally, the problem asks for $113l$. When we multiply our result by $113$, the denominator vanishes, leaving us with the elegant integer **1552**.

Enter Numerical Value:

Visualized Solution

Analyze Hyperbola $H$

Find Vertices of $H$

Relate $H$ to Ellipse $E$

Calculate Eccentricity $e_{H}$

Find Eccentricity $e_{E}$

Determine $a^{2}$ for Ellipse

Calculate Latus Rectum $l$

Final Answer

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

Analyzing the Setup

The Hyperbola's Hidden Orientation

The Geometric Bridge

The Eccentricity Dance

The Final Calculation

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Enter Numerical Value:

Visualized Solution

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

Find Vertices of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } H

Relate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } H to Ellipse .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E

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

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

Determine .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a2 for Ellipse

Calculate Latus Rectum .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } l