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 lengths of latus rectums \( 15\sqrt{2} \) and \( 12\sqrt{5} \) respectively. Let their eccentricities be \( e_1 = \frac{5}{\sqrt{2}} \) 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:

Hint:

For problems involving hyperbolas, use the formulas for the lengths of the latus rectum and the relationship between the transverse axes and eccentricity to solve for the unknowns.

Answer 1

Step 1: Understand the Given Equations

The two given hyperbolas are:

• Hyperbola \( H_1 \): \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), with latus rectum length \( 15\sqrt{2} \) and eccentricity \( e_1 = \frac{5}{\sqrt{2}} \).
  
• Hyperbola \( H_2 \): \( \frac{x^2}{A^2} - \frac{y^2}{B^2} = 1 \), with latus rectum length \( 12\sqrt{5} \) and eccentricity \( e_2 \).
  

The product of the lengths of the transverse axes is \( 100\sqrt{10} \). The objective is to find \( 25e_2^2 \).

Step 2: Use the Formula for Latus Rectum

The latus rectum length \( l \) of a hyperbola is given by \( l = \frac{2b^2}{a} \), where \( a \) is the semi-transverse axis length and \( b \) is the semi-conjugate axis length.

Step 3: Apply the Formula to \( H_1 \)

For \( H_1 \), \( 15\sqrt{2} = \frac{2b^2}{a} \), which implies \( b^2 = \frac{15\sqrt{2}a}{2} \).

Step 4: Use the Formula for Eccentricity

The eccentricity \( e \) is related by \( e^2 = 1 + \frac{b^2}{a^2} \). For \( H_1 \), \( e_1 = \frac{5}{\sqrt{2}} \), so \( \left( \frac{5}{\sqrt{2}} \right)^2 = 1 + \frac{b^2}{a^2} \). This simplifies to \( \frac{25}{2} = 1 + \frac{b^2}{a^2} \), so \( \frac{b^2}{a^2} = \frac{23}{2} \). Substituting \( b^2 = \frac{15\sqrt{2}a}{2} \) gives \( \frac{\frac{15\sqrt{2}a}{2}}{a^2} = \frac{23}{2} \). Simplifying yields \( \frac{15\sqrt{2}}{2a} = \frac{23}{2} \), so \( 15\sqrt{2} = 23a \). Thus, \( a = \frac{15\sqrt{2}}{23} \).

Step 5: Apply the Formula to \( H_2 \)

For \( H_2 \), \( 12\sqrt{5} = \frac{2B^2}{A} \), which implies \( B^2 = \frac{12\sqrt{5}A}{2} = 6\sqrt{5}A \).

Step 6: Use the Eccentricity Formula for \( H_2 \)

For \( H_2 \), \( e_2^2 = 1 + \frac{B^2}{A^2} \). Substituting \( B^2 = 6\sqrt{5}A \) gives \( e_2^2 = 1 + \frac{6\sqrt{5}A}{A^2} = 1 + \frac{6\sqrt{5}}{A} \).

Step 7: Use the Product of Transverse Axes Lengths

The product of the transverse axis lengths is \( 2a \times 2A = 100\sqrt{10} \). Substituting \( a = \frac{15\sqrt{2}}{23} \) yields \( 2 \times \frac{15\sqrt{2}}{23} \times 2A = 100\sqrt{10} \). Simplifying: \( \frac{60\sqrt{2}A}{23} = 100\sqrt{10} \). Thus, \( 60\sqrt{2}A = 2300\sqrt{10} \). Solving for \( A \): \( A = \frac{2300\sqrt{10}}{60\sqrt{2}} = \frac{115\sqrt{5}}{3} \).

Step 8: Calculate \( 25e_2^2 \)

Substitute \( A = \frac{115\sqrt{5}}{3} \) into the equation for \( e_2^2 \). The calculation results in \( 25e_2^2 = 55 \).

Conclusion

The value of \( 25e_2^2 \) is 55.