If f(x, y) = 0 is the combined equation of th | vedclass

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(A) The given curve is $3x^2 - 4xy + 5y^2 - 2x + y - 6 = 0$. The line is $4x - 6y - 2 = 0$,which implies $2x - 3y = 1$. To find the combined equation

Vedclass · Accepted Answer

$153$

Vedclass · Answer

(A) The given curve is $3x^2 - 4xy + 5y^2 - 2x + y - 6 = 0$. The line is $4x - 6y - 2 = 0$,which implies $2x - 3y = 1$. To find the combined equation of the lines joining the origin to the intersection points,we homogenize the curve using the line equation: $3x^2 - 4xy + 5y^2 - (2x - y)(2x - 3y) - 6(2x - 3y)^2 = 0$. Expanding this: $3x^2 - 4xy + 5y^2 - (4x^2 - 6xy - 2xy + 3y^2) - 6(4x^2 - 12xy + 9y^2) = 0$. $3x^2 - 4xy + 5y^2 - 4x^2 + 8xy - 3y^2 - 24x^2 + 72xy - 54y^2 = 0$. $f(x, y) = -25x^2 + 76xy - 52y^2 = 0$. Now,calculate $f(1, -1) = -25(1)^2 + 76(1)(-1) - 52(-1)^2 = -25 - 76 - 52 = -153$. Calculate $f(-1, -1) = -25(-1)^2 + 76(-1)(-1) - 52(-1)^2 = -25 + 76 - 52 = -1$. Therefore,$\frac{f(1, -1)}{f(-1, -1)} = \frac{-153}{-1} = 153$.

If $f(x, y) = 0$ is the combined equation of the lines joining the origin to the points where the line $4x - 6y - 2 = 0$ meets the curve $3x^2 - 4xy + 5y^2 - 2x + y - 6 = 0$,then $\frac{f(1, -1)}{f(-1, -1)} = $

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