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:
Show Hint
When calculating the area of a quadrilateral, use the coordinates of the vertices and the appropriate area formula.
Points of contact are provided as follows: \( A \left( \frac{-16}{5}, \frac{9}{5} \right), \quad B \left( \frac{16}{5}, \frac{-9}{5} \right) \), and point \( D \) is \( D \left( \frac{12}{5}, \frac{12}{5} \right). \)
Step 2: Triangle \( ABD \) Area:
The area of triangle \( ABD \) is calculated as: \[ \text{Area of } ABD = \frac{1}{2} \left| \begin{array}{ccc} \frac{-16}{5} & \frac{9}{5} & 1 \\ \frac{16}{5} & \frac{-9}{5} & 1 \\ \frac{12}{5} & \frac{12}{5} & 1 \\ \end{array} \right| = 12. \]
Step 3: Quadrilateral \( ABCD \) Area:
The area of quadrilateral \( ABCD \) is determined to be: \[ \text{Area of } ABCD = 24. \]
