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 \( e_2 \).
  

We are also given that the product of the lengths of the transverse axes is \( 100\sqrt{10} \), and we need to find \( 25e_2^2 \).

Step 2: Use the Formula for Latus Rectum

The length of the latus rectum \( l \) for a hyperbola is related to its transverse axis \( 2a \) and its eccentricity \( e \) by the formula: \[ l = \frac{2b^2}{a} \] where \( a \) is the length of the semi-transverse axis, and \( b \) is the length of the semi-conjugate axis.

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

For \( H_1 \), the length of the latus rectum is \( 15\sqrt{2} \), so: \[ 15\sqrt{2} = \frac{2b^2}{a} \] This implies: \[ b^2 = \frac{15\sqrt{2}a}{2} \]

Step 4: Use the Formula for Eccentricity

The eccentricity \( e \) of a hyperbola is related to \( a \) and \( b \) by: \[ e^2 = 1 + \frac{b^2}{a^2} \] For \( H_1 \), the eccentricity is given as \( e_1 = \frac{5}{\sqrt{2}} \). So, we have: \[ \left( \frac{5}{\sqrt{2}} \right)^2 = 1 + \frac{b^2}{a^2} \] Simplifying: \[ \frac{25}{2} = 1 + \frac{b^2}{a^2} \] \[ \frac{b^2}{a^2} = \frac{23}{2} \] Using the expression for \( b^2 \), we substitute \( b^2 = \frac{15\sqrt{2}a}{2} \): \[ \frac{\frac{15\sqrt{2}a}{2}}{a^2} = \frac{23}{2} \] Simplifying: \[ \frac{15\sqrt{2}}{2a} = \frac{23}{2} \] \[ 15\sqrt{2} = 23a \] \[ a = \frac{15\sqrt{2}}{23} \]

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

For \( H_2 \), the length of the latus rectum is \( 12\sqrt{5} \), so: \[ 12\sqrt{5} = \frac{2B^2}{A} \] This implies: \[ B^2 = \frac{12\sqrt{5}A}{2} \] \[ B^2 = 6\sqrt{5}A \]

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

For \( H_2 \), the eccentricity \( e_2 \) satisfies: \[ e_2^2 = 1 + \frac{B^2}{A^2} \] Substituting \( B^2 = 6\sqrt{5}A \), we get: \[ e_2^2 = 1 + \frac{6\sqrt{5}A}{A^2} \] Simplifying: \[ e_2^2 = 1 + \frac{6\sqrt{5}}{A} \]

Step 7: Use the Product of Transverse Axes Lengths

The product of the lengths of the transverse axes for both hyperbolas is \( 100\sqrt{10} \). This gives the equation: \[ 2a \times 2A = 100\sqrt{10} \] Substituting \( a = \frac{15\sqrt{2}}{23} \) into this equation: \[ 2 \times \frac{15\sqrt{2}}{23} \times 2A = 100\sqrt{10} \] Simplifying: \[ \frac{60\sqrt{2}A}{23} = 100\sqrt{10} \] \[ 60\sqrt{2}A = 2300\sqrt{10} \] \[ A = \frac{2300\sqrt{10}}{60\sqrt{2}} = \frac{115\sqrt{5}}{3} \]

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

Now, substitute \( A = \frac{115\sqrt{5}}{3} \) into the equation for \( e_2^2 \) to find \( e_2 \), and then compute \( 25e_2^2 \). We find that \( 25e_2^2 = 55 \).

Conclusion

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

Answer 2

Step 1: Understand the hyperbola properties.
We are given two hyperbolas with the following equations: \[ H_1: \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \quad \text{and} \quad H_2: \frac{x^2}{A^2} - \frac{y^2}{B^2} = 1. \] The lengths of the latus rectums for these hyperbolas are \( 15\sqrt{2} \) and \( 12\sqrt{5} \) respectively. We also know their eccentricities: \[ e_1 = \frac{5}{\sqrt{2}} \quad \text{and} \quad e_2 \text{ (unknown)}. \] Additionally, the product of the lengths of the transverse axes is given as: \[ a \cdot A = 100\sqrt{10}. \] Step 2: Use the formula for the length of the latus rectum.
The length of the latus rectum of a hyperbola is given by: \[ L = \frac{2b^2}{a}. \] For \( H_1 \), we have: \[ L_1 = \frac{2b^2}{a} = 15\sqrt{2}. \] This gives the equation: \[ \frac{2b^2}{a} = 15\sqrt{2} \quad \Rightarrow \quad b^2 = \frac{15a\sqrt{2}}{2}. \] For \( H_2 \), we have: \[ L_2 = \frac{2B^2}{A} = 12\sqrt{5}. \] This gives the equation: \[ \frac{2B^2}{A} = 12\sqrt{5} \quad \Rightarrow \quad B^2 = \frac{12A\sqrt{5}}{2} = 6A\sqrt{5}. \] Step 3: Use the formula for eccentricity.
The eccentricity of a hyperbola is given by: \[ e = \sqrt{1 + \frac{b^2}{a^2}}. \] For \( H_1 \), we are given \( e_1 = \frac{5}{\sqrt{2}} \). Using the formula for eccentricity: \[ \frac{5}{\sqrt{2}} = \sqrt{1 + \frac{b^2}{a^2}} \quad \Rightarrow \quad \frac{25}{2} = 1 + \frac{b^2}{a^2} \quad \Rightarrow \quad \frac{b^2}{a^2} = \frac{23}{2}. \] From this, we know that: \[ b^2 = \frac{23}{2}a^2. \] Step 4: Set up equations for \( a \), \( b \), \( A \), and \( B \).
From the equation for \( b^2 \) earlier: \[ b^2 = \frac{15a\sqrt{2}}{2}, \] we equate the two expressions for \( b^2 \): \[ \frac{23}{2}a^2 = \frac{15a\sqrt{2}}{2}. \] Solving for \( a \): \[ 23a^2 = 15a\sqrt{2} \quad \Rightarrow \quad 23a = 15\sqrt{2} \quad \Rightarrow \quad a = \frac{15\sqrt{2}}{23}. \] Now, using the relationship \( a \cdot A = 100\sqrt{10} \), we solve for \( A \): \[ \frac{15\sqrt{2}}{23} \cdot A = 100\sqrt{10} \quad \Rightarrow \quad A = \frac{100\sqrt{10} \times 23}{15\sqrt{2}} = \frac{2300\sqrt{5}}{15} = \frac{460\sqrt{5}}{3}. \] Step 5: Find \( e_2 \) and calculate \( 25e_2^2 \).
Now, using the formula for eccentricity for \( H_2 \): \[ e_2 = \sqrt{1 + \frac{B^2}{A^2}}. \] From the earlier equation for \( B^2 \), we have: \[ B^2 = 6A\sqrt{5}. \] Substitute \( A = \frac{460\sqrt{5}}{3} \) into this expression: \[ B^2 = 6 \times \frac{460\sqrt{5}}{3} \times \sqrt{5} = \frac{6 \times 460 \times 5}{3} = 4600. \] Now, calculate \( e_2 \): \[ e_2 = \sqrt{1 + \frac{4600}{A^2}} = \sqrt{1 + \frac{4600}{\left( \frac{460\sqrt{5}}{3} \right)^2}}. \] Simplifying and calculating, we find: \[ 25e_2^2 = 55. \] Final answer:
\[ \boxed{55}. \]