\( 8 \)
\( 12 \)
The three lines given are: 1. \( y = -4 \) (horizontal line at \( y = -4 \)) 2. \( y = x \) (diagonal line) 3. \( y = -4 \) (same as the first one) These lines form a right triangle with the x-axis. - The first and second lines intersect at the point \( (-4, -4) \).
- The second and third lines intersect at the point \( (4, 4) \).
Next, we calculate the area of the triangle. The base of the triangle is the distance between the x-axis and the line \( y = -4 \), which is 4 units. The height of the triangle is the distance between the line \( y = x \) and the x-axis, which is also 4 units.
Thus, the area of the triangle is: \[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 4 = 8 \] Therefore, the correct answer is (B) \( 8 \).
The center of a circle $ C $ is at the center of the ellipse $ E: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $, where $ a>b $. Let $ C $ pass through the foci $ F_1 $ and $ F_2 $ of $ E $ such that the circle $ C $ and the ellipse $ E $ intersect at four points. Let $ P $ be one of these four points. If the area of the triangle $ PF_1F_2 $ is 30 and the length of the major axis of $ E $ is 17, then the distance between the foci of $ E $ is: