Let H1:a2x2−b2y2=1 and H2:−A2x2+B2y2=1 be two hyperbolas having length of latus rectums 152 and 125 respectively. Let their eccentricities be e1=25 and e2 respectively. If the product of the lengths of their transverse axes is 10010, then 25e22 is equal to:
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Step-by-Step Solution
Step 1: Analyze Hyperbola H1
For the first hyperbola H1, the length of the latus rectum is given by L1=a2b2. We are given L1=152. The square of the eccentricity e12 is given by 1+a2b2. We are given e1=25, so e12=25.
Step 2: Solve for a and b for H1
From the eccentricity equation, we find a2b2=23, which means b2=23a2. Substitute this into the latus rectum equation to solve for a. Once a is found, we can find b2. The length of the transverse axis for H1 is 2a=102.
Step 3: Analyze Hyperbola H2
For the second hyperbola H2, the equation is −A2x2+B2y2=1, which means its transverse axis is along the y -axis. The length of the latus rectum is L2=B2A2, and we are given L2=125. The square of its eccentricity e22 is 1+B2A2.
Step 4: Solve for A and B for H2
We are given that the product of the lengths of their transverse axes is 10010. For H1, the transverse axis length is 2a=102. For H2, the transverse axis length is 2B. So, 2a⋅2B=10010. Using this, we can find B. Then, substitute B into the latus rectum equation for H2 to find A2.
Step 5: Calculate e22
Now that we have A2 and B2, we can calculate the square of the eccentricity e22 for the second hyperbola using the formula e22=1+B2A2.
Step 6: Calculate 25e22
Finally, we need to find the value of 25e22. Substitute the calculated value of e22 into this expression.