Question

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:

• A. 8
• B. 10
• C. 12
• D. 14

Hint:

For problems involving the foci of ellipses, use the relationship \( c = \sqrt{a^2 - b^2} \) and apply geometric properties such as the area of a triangle to solve for unknowns.

Answer 1

The Correct Option is B

The equation of the ellipse is given as \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, \quad a>b \] - The foci \( F_1 \) and \( F_2 \) are at \( (\pm c, 0) \), where \( c = \sqrt{a^2 - b^2} \).
- The length of the major axis is \( 2a = 17 \), resulting in \( a = \frac{17}{2} = 8.5 \).
- The area of triangle \( PF_1F_2 \) is 30. The area of this triangle is calculated as \( \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2c \times b = cb \). Given the area is 30, we have \( cb = 30 \).
- We can substitute \( b \) using \( b = \sqrt{a^2 - c^2} \). From the equation for \( c \), \( b = \sqrt{a^2 - (a^2 - b^2)} = \sqrt{b^2} \).
- Solving for \( 2c \), the distance between the foci, yields \( c = 5 \), and thus \( 2c = 10 \). The correct answer is \( 10 \).