Let H₁: x²/a² - y²/b² = 1 and H₂: -x²/A² + y²/B² = 1 be two hyperbolas having length of latus rectum

Q: Wait, why is the latus rectum formula $\frac{2A^2}{B}$ for $H_2$ instead of $\frac{2b^2}{a}$?

Great observation! The formula for the latus rectum is always $\frac{2 \times (\text{semi-conjugate axis})^2}{\text{semi-transverse axis}}$. For a horizontal hyperbola, the transverse axis is $2a$ and the conjugate is $2b$, so $LR = \frac{2b^2}{a}$. For a vertical hyperbola, the transverse axis is $2B$ and the conjugate is $2A$, so $LR = \frac{2A^2}{B}$. Always identify the transverse axis first!

Q: Why can't we just use $c^2 = a^2 + b^2$ for both hyperbolas?

You absolutely can! The relation $e^2 = 1 + \frac{b^2}{a^2}$ is derived directly from $c^2 = a^2 + b^2$ and $e = \frac{c}{a}$. Using the eccentricity formula is often faster in competitive exams because it directly relates the given $e$ to the ratio of the axes, saving you from calculating the focal distance $c$ unnecessarily.

Q: I got $A^2 = 150$ and $B^2 = 125$. Did I make a mistake?

Not at all! You are on the right track. $B = 5\sqrt{5}$, so $B^2 = 25 \times 5 = 125$. And $A^2 = 150$ is correct. Your calculation of $e_2^2 = 1 + \frac{150}{125} = 1 + \frac{6}{5} = \frac{11}{5}$ is perfectly accurate. Keep going!

Question

Let $H_1: \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ and $H_2: -\frac{x^{2}}{A^{2}}+\frac{y^{2}}{B^{2}}=1$ be two hyperbolas having length of latus rectums $15\sqrt{2}$ and $12\sqrt{5}$ respectively. Let their ecentricities be $e_{1}=\sqrt{\frac{5}{2}}$ and $e_2$ respectively. If the product of the lengths of their transverse axes is $100\sqrt{10},$ then $25e_{2}^{2}$ is equal to

Sigma Percentile · Accepted Answer

### The Geometry of Duality: Mastering Hyperbolas

Welcome, future engineers. Today, we are going to dissect a problem that tests not just your algebraic manipulation, but your ability to maintain clarity amidst the duality of two distinct hyperbolas.

In the world of JEE Advanced, the hyperbola is often the 'rebel' of the conic sections. While the ellipse is closed and contained, the hyperbola is open, reaching out toward infinity. Let us tame these two curves.

### Phase 1: Decoding the Horizontal Rebel ($H_1$)

We begin with $H_1: \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. This is our classic horizontal hyperbola. We are given the eccentricity $e_1 = \sqrt{\frac{5}{2}}$.

Recall the fundamental identity for a horizontal hyperbola:
$$e_1^2 = 1 + \frac{b^2}{a^2}$$

Substituting our value, we get $\frac{5}{2} = 1 + \frac{b^2}{a^2}$, which simplifies to $\frac{b^2}{a^2} = \frac{3}{2}$, or $b^2 = \frac{3}{2}a^2$. This is our first vital link.

Next, we look at the latus rectum, $LR_1 = 15\sqrt{2}$. The formula for the latus rectum of a horizontal hyperbola is $LR_1 = \frac{2b^2}{a}$.

By substituting our expression for $b^2$, we get:
$$\frac{2(\frac{3}{2}a^2)}{a} = 15\sqrt{2}$$

The algebra here is elegant: the $2$ cancels, one $a$ cancels, and we are left with $3a = 15\sqrt{2}$, which gives us $a = 5\sqrt{2}$. We have successfully pinned down the first parameter of our first hyperbola.

### Phase 2: The Bridge of Transverse Axes

Now, we connect the two worlds. The problem states that the product of the lengths of their transverse axes is $100\sqrt{10}$.

For $H_1$, the transverse axis is $2a$. For $H_2$, which is a vertical hyperbola, the transverse axis lies along the y-axis and has a length of $2B$. Thus:
$$2a 	imes 2B = 100\sqrt{10}$$

Since we know $a = 5\sqrt{2}$, then $2a = 10\sqrt{2}$. Substituting this, we have $10\sqrt{2} 	imes 2B = 100\sqrt{10}$, which simplifies to $20\sqrt{2}B = 100\sqrt{10}$.

Dividing both sides by $20\sqrt{2}$, we find $B = 5\sqrt{5}$. The bridge is built.

### Phase 3: Unveiling the Vertical Rebel ($H_2$)

Finally, we turn to $H_2: -\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$. This is a vertical hyperbola. We are given $LR_2 = 12\sqrt{5}$.

For this orientation, the latus rectum formula is $LR_2 = \frac{2A^2}{B}$. We know $B = 5\sqrt{5}$, so:
$$\frac{2A^2}{5\sqrt{5}} = 12\sqrt{5}$$

Multiplying both sides by $5\sqrt{5}$, we get $2A^2 = 12 	imes 5 	imes 5 = 300$, which means $A^2 = 150$.

We are at the finish line. We need $25e_2^2$. For a vertical hyperbola, the eccentricity relation is:
$$e_2^2 = 1 + \frac{A^2}{B^2}$$

We have $A^2 = 150$ and $B^2 = (5\sqrt{5})^2 = 125$. Thus:
$$e_2^2 = 1 + \frac{150}{125} = 1 + \frac{6}{5} = \frac{11}{5}$$

The final calculation is $25 	imes \frac{11}{5} = 5 	imes 11 = 55$.

And there it is—the result of our journey. Remember, in JEE Advanced, the math is never just about the numbers; it is about the symmetry and the relationships between them. You have mastered the hyperbolas today. The final answer is **55**.

Enter Numerical Value:

Visualized Solution

Visualizing the Hyperbolas

Eccentricity Formula for $H_{1}$

Substituting $e_{1}$

Solving for $b^{2}$

Latus Rectum of $H_{1}$

Finding Parameter $a$

Product of Transverse Axes

Solving for Parameter $B$

Latus Rectum of $H_{2}$

Finding Parameter $A^{2}$

Eccentricity Formula for $H_{2}$

Calculating $e_{2}^{2}$

Final Calculation

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

The Geometry of Duality

Phase 1

Phase 2

Phase 3

Similar Questions

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Enter Numerical Value:

Visualized Solution

Visualizing the Hyperbolas

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