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

We are given the equation of the ellipse: \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, \quad a>b \] - The foci \( F_1 \) and \( F_2 \) of the ellipse are located at \( (\pm c, 0) \), where \( c \) is given by: \[ c = \sqrt{a^2 - b^2} \] - The length of the major axis of the ellipse is \( 2a = 17 \), so: \[ a = \frac{17}{2} = 8.5 \] - The area of the triangle \( PF_1F_2 \) is given as 30. The area of a triangle with base \( 2c \) and height \( b \) (since the height of the triangle is the distance from the point \( P \) to the major axis, which is the semi-minor axis of the ellipse) is given by: \[ \text{Area of triangle} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2c \times b = cb \] Given that the area is 30: \[ cb = 30 \] - Now, we can substitute the value of \( b \) (the semi-minor axis) using the relation \( b = \sqrt{a^2 - c^2} \). From the earlier equation for \( c \), we have: \[ b = \sqrt{a^2 - (a^2 - b^2)} = \sqrt{b^2} \]
Thus, we can solve for the distance between the foci, \( 2c \). The solution is given by: \[ c = 5, \quad \text{so the distance between the foci is } 2c = 10 \]
Thus, the correct answer is \( 10 \).