If the points of intersection of two distinct conics x2+y2=4b and x216+ y2b2=1 lie on the curve y2=3x2, then 3 3 times the area of the rectangle formed by the intersection points is .
Answer
432
Solution
Put y2=3x2 in the circle: 4x2=4b x2=b
In the ellipse: b16+ 3bb2=1 b2-16b+48=0 b=4 or 12
b = 4 gives the same circle x2+y2=16 twice (not distinct) ⇒ b = 12
Points: x2=12, y2=36 ⇒ ( 2 3, 6 )
Rectangle area =4 3 12=48 3 ⇒ 3 3 48 3=432
Answer: 432