Let C be the circle of minimum area enclosing the ellipse E : x2a2 + y2b2 = 1 with eccentricity 12 and foci ( 2, 0). Let PQR be a variable triangle, whose vertex P is on the circle C and the side QR of length 2a is parallel to the major axis of E and contains the point of intersection of E with the negative y-axis. Then the maximum area of the triangle PQR is :
- 8(3 + 2)
- 8(2 + 3)
- 6(3 + 2)
- 6(2 + 3)
Answer
(2) 8(2 + 3)
Solution
Foci ( ae, 0) = ( 2, 0) and e = 12
a = 4, b2 = a2(1 - e2) = 16 34 = 12 b = 2 3
The smallest circle enclosing E is its auxiliary circle C : x2 + y2 = 16 (radius 4).
E meets the negative y-axis at (0, -2 3), so QR lies on the line y = -2 3 and QR = 2a = 8.
Area is maximum when P is the point of C farthest from this line, i.e. P(0, 4):
height = 4 + 2 3
maximum area = 12 8 (4 + 2 3)
= 16 + 8 3 = 8(2 + 3)
Answer: (2)