Question

Given that \( a \) and \( b \) are the semi-major and semi-minor axes of an ellipse whose axes are along the coordinate axes. If its latus rectum is of length 4 units and the distance between its foci is \( 4\sqrt{2} \), then the value of \( a^2 + b^2 \) is:

• A. \( 24 \)
• B. \( 18 \)
• C. \( 16 \)
• D. \( 12 \)

Hint:

For problems involving the latus rectum of an ellipse, use the formula \( \frac{2b^2}{a} \). Also, use \( c^2 = a^2 - b^2 \) to find missing parameters.

Answer 1

The Correct Option is A

Step 1: Standard form of the ellipse 
The equation of an ellipse centered at the origin with its major axis along the x-axis is: \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1. \] The foci of the ellipse are located at: \[ (\pm c, 0), \] where \( c^2 = a^2 - b^2 \). 

Step 2: Use the given foci distance 
The total distance between the foci is given as \( 4\sqrt{2} \), so: \[ 2c = 4\sqrt{2}. \] Solving for \( c \): \[ c = 2\sqrt{2}. \] 

Step 3: Use the latus rectum formula 
The length of the latus rectum of an ellipse is given by: \[ \frac{2b^2}{a}. \] It is given that the latus rectum is 4, so: \[ \frac{2b^2}{a} = 4. \] Solving for \( b^2 \): \[ b^2 = 2a. \] 

Step 4: Solve for \( a^2 + b^2 \) 
From the foci relation: \[ c^2 = a^2 - b^2. \] Substituting \( c = 2\sqrt{2} \): \[ (2\sqrt{2})^2 = a^2 - b^2. \] \[ 8 = a^2 - b^2. \] Using \( b^2 = 2a \), substitute: \[ a^2 - 2a = 8. \] Rearrange: \[ a^2 - 2a - 8 = 0. \] Factoring: \[ (a - 4)(a + 2) = 0. \] Since \( a>0 \), we take \( a = 4 \). Substituting into \( b^2 = 2a \): \[ b^2 = 2(4) = 8. \] Thus: \[ a^2 + b^2 = 16 + 8 = 24. \] 

Step 5: Conclusion 
Thus, the correct answer is: \[ \mathbf{24}. \]