Mathematics

Mathematics

Problem

Transform the equation x2−3xy+11x−12y+36=0x^2 - 3xy + 11x - 12y + 36 = 0 to parallel axes through the point (−4,1)(-4, 1) becomes ax2+bxy+1=0ax^2 + bxy + 1 = 0 then b2−a=b^2 - a = a. 14\frac{1}{4} b. 116\frac{1}{16} c. 164\frac{1}{64} d. 1256\frac{1}{256}
Intuitive explanation

When you shift a coordinate system to a new origin, you're essentially substituting x=X−4x = X - 4 and y=Y+1y = Y + 1 (where capital letters are the new coordinates). The key insight is that this translation eliminates the linear terms and reveals the "pure" quadratic structure of the curve. After substitution and simplification, you'll get an equation in XX and YY with no first-degree terms—compare its coefficients to the target form aX2+bXY+1=0aX^2 + bXY + 1 = 0 to find aa and bb, then compute b2−ab^2 - a.

Intuitive explanation

When you shift a coordinate system to a new origin, you're essentially substituting x=X−4x = X - 4 and y=Y+1y = Y + 1 (where capital letters are the new coordinates). The key insight is that this translation eliminates the linear terms and reveals the "pure" quadratic structure of the curve. After substitution and simplification, you'll get an equation in XX and YY with no first-degree terms—compare its coefficients to the target form aX2+bXY+1=0aX^2 + bXY + 1 = 0 to find aa and bb, then compute b2−ab^2 - a.

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