Question

If the extremities of the latus recta having positive ordinate of the ellipse $$ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \quad (a > b) $$ lie on the parabola $$ x^2 + 2ay - 4 = 0, $$ then the points $(a, b)$ lie on the curve:

• A. \( xy = 4 \)
• B. \( x^2 + y^2 = 4 \)
• C. \( \frac{x^2}{4} + \frac{y^2}{4} = 1 \)
• D. \( \frac{x^2}{4} - \frac{y^2}{4} = 1 \)

Hint:

For problems involving latus rectum coordinates in conic sections, use the standard form \( ( \pm c, \frac{b^2}{a} ) \) and substitute into the given curve equation to find the relationship.

Answer 1

The Correct Option is B

Step 1: Find the coordinates of the extremities of the latus rectum 
The latus rectum of an ellipse is given by the coordinates: \[ \left( \pm c, \frac{b^2}{a} \right), \] where \( c \) is the focal distance: \[ c = \sqrt{a^2 - b^2}. \] Thus, the coordinates of the extremities of the latus rectum with positive ordinate are: \[ \left( c, \frac{b^2}{a} \right) \quad \text{and} \quad \left( -c, \frac{b^2}{a} \right). \] 

Step 2: Use the given parabola equation 
The extremities satisfy the equation of the given parabola: \[ x^2 + 2ay - 4 = 0. \] Substituting \( x = c = \sqrt{a^2 - b^2} \) and \( y = \frac{b^2}{a} \): \[ (\sqrt{a^2 - b^2})^2 + 2a \left( \frac{b^2}{a} \right) - 4 = 0. \] \[ a^2 - b^2 + 2b^2 - 4 = 0. \] \[ a^2 + b^2 - 4 = 0. \] \[ a^2 + b^2 = 4. \] 

Step 3: Conclusion 
Thus, the points \( (a, b) \) satisfy: \[ x^2 + y^2 = 4. \] Thus, the correct answer is: \[ \mathbf{x^2 + y^2 = 4}. \]