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C 1 equiv x^ 2 y^ 2 -169=0 anc C 2 =x^ 2 y^ 2 -24x-32y 111=0 0. Let AB be their C^ prime common chord. Maximum area of Delta*A * B * C , if ' on minor arc AB, is equal to 12sqrt(a) Find greate prime divisor of a Let?
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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:

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C 1 equiv x^ 2 y^ 2 -169=0 anc C 2 =x^ 2 y^ 2 -24x-32y 111=0 0. Let AB be their C^ prime common chord. Maximum area of Delta*A * B * C , if ' on minor arc AB, is equal to 12sqrt(a) Find greate prime divisor of a Let?
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