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

Step 1: Given points of contact: 

\[ A \left( \frac{-16}{5}, \frac{9}{5} \right), \quad B \left( \frac{16}{5}, \frac{-9}{5} \right) \] and the point \( D \) is: \[ D \left( \frac{12}{5}, \frac{12}{5} \right). \]

Step 2: Area Calculation of Triangle \( ABD \):

The area of triangle \( ABD \) is given by: \[ \text{Area of } ABD = \frac{1}{2} \left| \begin{array}{ccc} \frac{-16}{5} & \frac{9}{5} & 1 \\ \frac{16}{5} & \frac{-9}{5} & 1 \\ \frac{12}{5} & \frac{12}{5} & 1 \\ \end{array} \right| = 12. \]

Step 3: Area of Quadrilateral \( ABCD \):

The area of quadrilateral \( ABCD \) is: \[ \text{Area of } ABCD = 24. \]