C 1 equiv x^ 2 y^ 2 -169=0 anc C 2 =x^ 2 y^ 2 -24x-32y 111=0 0. Let ...
Problem:
Given two curves C1: x^2y^2 - 169 = 0 and C2: x^2y^2 - 24x - 32y + 111 = 0. Let AB be their common chord. The maximum area of DeltaABC, where C lies on the minor arc AB, is equal to 12sqrt(a). Find the greatest prime divisor of a.
Solution:
Finding the Common Chord AB
To find the common chord AB, we need to solve the system of equations:
C1: x^2y^2 - 169 = 0
C2: x^2y^2 - 24x - 32y + 111 = 0
Subtracting the second equation from the first, we get:
24x + 32y - 111 = 0
Dividing by 8, we get:
3x + 4y - 13.875 = 0
This is the equation of the common chord AB.
Finding the Coordinates of A and B
To find the coordinates of A and B, we need to substitute the equation of the common chord AB into the equations of the curves C1 and C2.
For C1:
x^2y^2 - 169 = 0
3x + 4y - 13.875 = 0
Substituting 3x + 4y - 13.875 for x^2y^2 in the first equation, we get:
(3x + 4y - 13.875)^2 - 169 = 0
Expanding the square, we get:
9x^2 + 16y^2 + 2xy - 25.5x - 36y + 120.515625 - 169 = 0
Simplifying, we get:
9x^2 + 16y^2 + 2xy - 25.5x - 36y - 48.484375 = 0
For C2:
x^2y^2 - 24x - 32y + 111 = 0
3x + 4y - 13.875 = 0
Substituting 3x + 4y - 13.875 for x^2y^2 in the first equation, we get:
(3x + 4y - 13.875)^2 - 24x - 32y + 111 = 0
Expanding the square, we get:
9x^2 + 16y^2 + 2xy - 25.5x - 36y + 120.515625 - 24x - 32y + 111 = 0
Simplifying, we get: