For the first hyperbola H_1, the length of the latus rectum is given by L_1 = (2b²)/(a). We are given L_1 = 15√2. The square of the eccentricity e_1² is given by 1 + (b²)/(a²). We are given e_1 = √((5)/(2)), so e_1² = (5)/(2).

For the first hyperbola H_1, the length of the latus rectum is given by L_1 = (2b²)/(a). We are given L_1 = 15√2. The square of the eccentricity e_1² is given by 1 + (b²)/(a²). We are given e_1 = √((5)/(2)), so e_1² = (5)/(2).

We are given that the product of the lengths of their transverse axes is 100√10. For H_1, the transverse axis length is 2a = 10√2. For H_2, the transverse axis length is 2B. So, 2a · 2B = 100√10. Using this, we can find B. Then, substitute B into the latus rectum equation for H_2 to find A².

We are given that the product of the lengths of their transverse axes is 100√10. For H_1, the transverse axis length is 2a = 10√2. For H_2, the transverse axis length is 2B. So, 2a · 2B = 100√10. Using this, we can find B. Then, substitute B into the latus rectum equation for H_2 to find A².

Now that we have A² and B², we can calculate the square of the eccentricity e_2² for the second hyperbola using the formula e_2² = 1 + (A²)/(B²).

Finally, we need to find the value of 25e_2². Substitute the calculated value of e_2² into this expression.

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 eccentricities 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 $25 e_2^2$ is equal to:

Answer: 55

Step-by-step solution

Step 1: Analyze Hyperbola H1

For the first hyperbola $H_1$ , the length of the latus rectum is given by $L_1 = \frac{2b^2}{a}$ . We are given $L_1 = 15\sqrt{2}$ . The square of the eccentricity $e_1^2$ is given by $1 + \frac{b^2}{a^2}$ . We are given $e_1 = \sqrt{\frac{5}{2}}$ , so $e_1^2 = \frac{5}{2}$ .

Step 2: Solve for a and b for H1

From the eccentricity equation, we find $\frac{b^2}{a^2} = \frac{3}{2}$ , which means $b^2 = \frac{3}{2}a^2$ . Substitute this into the latus rectum equation to solve for $a$ . Once $a$ is found, we can find $b^2$ . The length of the transverse axis for $H_1$ is $2a = 10\sqrt{2}$ .

Step 3: Analyze Hyperbola H2

For the second hyperbola $H_2$ , the equation is $-\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$ , which means its transverse axis is along the $y$ -axis. The length of the latus rectum is $L_2 = \frac{2A^2}{B}$ , and we are given $L_2 = 12\sqrt{5}$ . The square of its eccentricity $e_2^2$ is $1 + \frac{A^2}{B^2}$ .

Step 4: Solve for A and B for H2

We are given that the product of the lengths of their transverse axes is $100\sqrt{10}$ . For $H_1$ , the transverse axis length is $2a = 10\sqrt{2}$ . For $H_2$ , the transverse axis length is $2B$ . So, $2a \cdot 2B = 100\sqrt{10}$ . Using this, we can find $B$ . Then, substitute $B$ into the latus rectum equation for $H_2$ to find $A^2$ .

Step 5: Calculate $e_2^2$

Now that we have $A^2$ and $B^2$ , we can calculate the square of the eccentricity $e_2^2$ for the second hyperbola using the formula $e_2^2 = 1 + \frac{A^2}{B^2}$ .

Step 6: Calculate $25e_2^2$

Finally, we need to find the value of $25e_2^2$ . Substitute the calculated value of $e_2^2$ into this expression.

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