The two orthogonal parabolas:
y=2x^2+a
x= 2y^2+b
intersect at four distinct points.
These four points are on the same circle of area 10.
Given that a, b < 0, determine the minimum value of ab.
If the circle has area 10, then its equation, for some {p,q} is p^2+q^2=10/pi.
Note: three of the 4 distinct points (if 4 exist) will always lie on some three-point circle, c. By symmetry, two of the points will be on a diameter of c, and the 4th point will be a reflection of one of those points about that diameter and thus will also be on c.
I observe that for such a pair of orthogonal parabolas, the centre of circle c is at (1/(2r),(1/2s) where r, s are the coefficients of the squared terms. Proof is left to the reader.
In this case, the centre of the circle will therefore be at the point (1/4,1/4), so that p=(x+1/4), and q=(y+1/4).
Now we have (x+1/4)^2+(y+1/4)^2=10/pi, y=(2x^2+a-1/2), x= (2y^2+b-1/2), with 4 solutions. One value, say a, is always 1/8-10/pi, while b has the same value, a larger negative value (discarded) and a smaller positive value (forbidden by the terms of the problem).
Thus ab=(1/8-10/pi)^2, around 9.3519.
Edited on March 14, 2025, 2:52 am
|
|
Posted by broll
on 2025-03-14 01:47:23 |
Possible approach.
