Let x²/a² + y²/b² = 1, a b be an ellipse, whose eccentricity is 1/√2 and the length of the latus r

Q: Wait, why do we ignore the Latus Rectum value of $\sqrt{14}$? Isn't it important?

That is a classic JEE trap! The problem asks for the square of the eccentricity, which depends only on the ratio of the semi-axes, not their absolute values. The Latus Rectum is provided to tempt you into calculating $a$ and $b$ individually, which is a waste of precious exam time.

Q: Why is the formula for the hyperbola's eccentricity different from the ellipse's?

Great observation! For an ellipse, the sum of distances to the foci is constant, leading to $b^2 = a^2(1-e^2)$. For a hyperbola, the difference is constant, leading to $b^2 = a^2(e^2-1)$. This sign difference is the geometric soul of the distinction between these two conic sections.

Q: What if the hyperbola equation had an 'm' coefficient, like $\frac{x^2}{a^2} - \frac{my^2}{b^2} = 1$?

If there were an 'm', you would rewrite it as $\frac{x^2}{a^2} - \frac{y^2}{(b^2/m)} = 1$. The new $b^2$ would be $b^2/m$. You would then use the ratio $\frac{b^2}{ma^2}$ in the eccentricity formula. Always reduce to the standard form first!

Question

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

Options:

(A) 3

(B) 7/2

(C) 3/2

(D) 5/2

Sigma Percentile · Accepted Answer

### The Elegance of Conic Ratios

Welcome, fellow traveler on the path to JEE mastery. Today, we are going to dissect a problem that, at first glance, seems to demand a mountain of calculation.

But as we peel back the layers, we will find that the true path lies in understanding the geometric soul of conic sections. Let us begin.

### Phase 1: The Ellipse and the Hidden Ratio

We start with the standard equation of an ellipse:

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

where $a > b$. We are given the eccentricity $e = \frac{1}{\sqrt{2}}$.

Now, pause for a moment. The eccentricity is the measure of how 'squashed' the ellipse is. The formula connecting this to the axes is:

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

Let us substitute our given value:

$$ \left(\frac{1}{\sqrt{2}}\right)^2 = 1 - \frac{b^2}{a^2} $$

This simplifies beautifully to $\frac{1}{2} = 1 - \frac{b^2}{a^2}$. Rearranging this, we find that:

$$ \frac{b^2}{a^2} = \frac{1}{2} $$

This ratio is the key to the entire problem. It is the DNA of our ellipse.

### Phase 2: The Latus Rectum Trap

Now, here is where many students stumble. The problem provides the length of the latus rectum as $\sqrt{14}$.

You might feel an urge to use the formula $L.R. = \frac{2b^2}{a} = \sqrt{14}$ to solve for $a$ and $b$. But stop!

Ask yourself: does the final question ask for $a$ or $b$? No. It asks for the square of the eccentricity of a hyperbola.

We have already extracted the only piece of information we need: the ratio $\frac{b^2}{a^2} = \frac{1}{2}$. The latus rectum is a distraction, a siren song designed to lead you into a swamp of unnecessary algebra. Ignore it and move forward with confidence.

### Phase 3: The Hyperbola Connection

We now turn our attention to the hyperbola:

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

We need to find the square of its eccentricity, $e_H^2$. The formula for the eccentricity of a hyperbola is:

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

Notice the elegance here—the sign has flipped from the ellipse formula. This is the fundamental difference between the two curves.

Since we already know that $\frac{b^2}{a^2} = \frac{1}{2}$, we simply plug this into our hyperbola formula:

$$ e_H^2 = 1 + \frac{1}{2} $$

### The Final Celebration

With a simple addition, we arrive at:

$$ e_H^2 = \frac{3}{2} $$

That is it. We have navigated the trap, identified the core ratio, and arrived at the solution with minimal effort.

This is the beauty of JEE Advanced mathematics—it rewards those who look for the underlying structure rather than those who blindly calculate. Keep this mindset, and you will find that even the most intimidating problems become elegant puzzles waiting to be solved. The final answer is **$\frac{3}{2}$**.

Select Answer:

Visualized Solution

Given Ellipse Parameters

Eccentricity Formula for Ellipse

Substituting the Eccentricity

Solving for the Ratio $\frac{b ^{2}}{a ^{2}}$

The Latus Rectum Trap

Introducing the Hyperbola

Eccentricity Formula for Hyperbola

Substituting the Ratio

Final Calculation

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

The Elegance of Conic Ratios

Phase 1

Phase 2

Phase 3

The Final Celebration

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