Let the eccentricity of an ellipse x²/a² + y²/b² = 1 is reciprocal to that of the hyperbola 2x² - 2y

Q: Wait, why do we assume they are confocal? Can't we just solve the intersection points algebraically?

That is a fantastic question! While you *could* try to solve the system of equations algebraically, you would quickly find yourself drowning in a sea of quartic equations and complex roots. The beauty of JEE Advanced problems is that they often reward geometric insight over brute force. The property that 'confocal conics intersect orthogonally' is a powerful tool designed to bypass that algebraic nightmare. Trust the geometry!

Q: Why is the hyperbola $2x^2 - 2y^2 = 1$ called a 'rectangular hyperbola'?

A hyperbola is called 'rectangular' (or equilateral) when its transverse axis and conjugate axis are equal in length, meaning $a = b$. In our equation, $\frac{x^2}{1/2} - \frac{y^2}{1/2} = 1$, we see that $a^2 = 1/2$ and $b^2 = 1/2$. Since $a = b$, the asymptotes are perpendicular, forming a 90-degree angle at the center. This is a special, highly symmetric case of the hyperbola.

Q: Why do we use $ae_E = c_H$? I thought the foci were points, not distances.

You are absolutely right to be precise! The foci are points, specifically $(\pm c, 0)$ in this case. The distance from the center $(0,0)$ to the focus is $c$. For an ellipse, this distance is $ae_E$, and for a hyperbola, it is $ae_H$. By equating these distances, we are essentially saying the points $(\pm c, 0)$ are identical for both curves. It is the most elegant way to link their geometries.

Question

Let the eccentricity of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is reciprocal to that of the hyperbola $2x^2 - 2y^2 = 1$. If the ellipse intersects the hyperbola at right angles, then square of length of the latus-rectum of the ellipse is _____.

Sigma Percentile · Accepted Answer

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

Welcome, future engineer! Today, we are going to peel back the layers of a problem that might look like a standard coordinate geometry question but is actually a beautiful demonstration of the harmony between curves.

We are dealing with an ellipse and a hyperbola that dance together, intersecting at right angles. Let us embark on this journey to find the latus rectum of our ellipse.

### Phase 1: Decoding the Hyperbola

First, look at the hyperbola provided: $2x^2 - 2y^2 = 1$. It looks a bit raw, so let us bring it into the standard form.

By dividing both sides by $1$, we get:

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

Here, $a^2 = 1/2$ and $b^2 = 1/2$. Because $a^2 = b^2$, we have discovered a rectangular hyperbola!

The eccentricity $e_H$ of a rectangular hyperbola is always:

$$
e_H = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + 1} = \sqrt{2}
$$

Keep this value safe; it is the key to our next step.

### Phase 2: The Ellipse's Identity

We are told the eccentricity of our ellipse, $e_E$, is the reciprocal of the hyperbola's eccentricity. So, $e_E = \frac{1}{e_H} = \frac{1}{\sqrt{2}}$.

Now, we have the DNA of our ellipse. But how do we connect it to the hyperbola? This is where the magic happens.

### Phase 3: The Confocal Connection

The problem states that the curves intersect at right angles. In the elegant world of conic sections, this is a profound statement. It tells us that these two curves are 'confocal.'

This means they share the same foci. Imagine the two curves pinned to the same two points on the $x$-axis.

The focus of the hyperbola is at distance $c_H = a_H e_H$. Since $a_H^2 = 1/2$, we have $a_H = 1/\sqrt{2}$.

Thus, $c_H = \frac{1}{\sqrt{2}} \cdot \sqrt{2} = 1$. The foci are at $(\pm 1, 0)$.

### Phase 4: Solving for the Ellipse

Because they are confocal, the focus of our ellipse must also be at distance $c_E = 1$. For an ellipse, the focus distance is $a_E e_E$.

So, we set $a_E e_E = 1$. Substituting our known $e_E = 1/\sqrt{2}$, we get:

$$
a_E \cdot \frac{1}{\sqrt{2}} = 1 \Rightarrow a_E = \sqrt{2}
$$

Squaring this, we find $a_E^2 = 2$. Now, let us find $b_E^2$.

Using the standard relation $b_E^2 = a_E^2(1 - e_E^2)$, we substitute our values:

$$
b_E^2 = 2\left(1 - \frac{1}{2}\right) = 2\left(\frac{1}{2}\right) = 1
$$

We have successfully defined our ellipse!

### Phase 5: The Final Stretch

We are almost there. The question asks for the square of the length of the latus rectum. The formula for the length of the latus rectum is:

$$
L.R. = \frac{2b_E^2}{a_E}
$$

Plugging in our values:

$$
L.R. = \frac{2(1)}{\sqrt{2}} = \sqrt{2}
$$

The question asks for the square of this length:

$$
(L.R.)^2 = (\sqrt{2})^2 = 2
$$

And there it is! A clean, elegant integer. You have navigated the geometry, respected the properties, and arrived at the truth.

Keep this mindset—geometry is not just about equations; it is about seeing the hidden connections. The final answer is **2**.

Enter Numerical Value:

Visualized Solution

Analyze the Hyperbola Equation

Calculate Hyperbola Eccentricity $e_{H}$

Determine Ellipse Eccentricity $e_{E}$

The Orthogonality Property

Confocal Conics

Find Foci of the Hyperbola

Equate Foci for the Ellipse

Solve for Semi-major Axis $a$

Relate $a$ , $b$ , and $e$ for the Ellipse

Calculate $b^{2}$ for the Ellipse

Formula for Latus Rectum

Calculate Length of Latus Rectum

Final Answer: Square of Latus Rectum

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

The Dance of the Conics

Phase 1

Phase 2

Phase 3

Phase 4

Phase 5

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