Question

Let for two distinct values of $ p $, the lines $ y = x + p $ touch the ellipse $ E: \frac{x^2}{4} + \frac{y^2}{9} = 1 $ at the points $ A $ and $ B $. Let the line $ y = x $ intersect $ E $ at the points $ C $ and $ D $. Then the area of the quadrilateral $ ABCD $ is equal to:

• A. 36
• B. 24
• C. 48
• D. 20

Hint:

When calculating the area of a quadrilateral, use the coordinates of the vertices and the appropriate area formula.

Answer 1

The Correct Option is B

We are given that the lines \( y = x + p \) are tangents to the ellipse \( E \) at points \( A \) and \( B \), and the line \( y = x \) intersects the ellipse at points \( C \) and \( D \). 
After finding the coordinates of the points \( A \), \( B \), \( C \), and \( D \), 
we use the formula for the area of a quadrilateral formed by these points to calculate the area. 
The result is 24. 
Thus, the correct answer is \( 24 \).