If the line x + alpha y + beta = 0 touches th | vedclass

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(C) The given equation is $4x^3 + 4y^3 = xy(xy + 16)$.Rearranging the terms,we get $4x^3 + 4y^3 - x^2y^2 - 16xy = 0$.Factoring the expression,we obtai

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$2$

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(C) The given equation is $4x^3 + 4y^3 = xy(xy + 16)$.Rearranging the terms,we get $4x^3 + 4y^3 - x^2y^2 - 16xy = 0$.Factoring the expression,we obtain $(4x - y^2)(x^2 - 4y) = 0$.This represents the combined equation of two parabolas: $y^2 = 4x$ and $x^2 = 4y$.The line $x + \alpha y + \beta = 0$ is a common tangent to these two parabolas.The common tangent to $y^2 = 4x$ and $x^2 = 4y$ is $x + y + 1 = 0$.Comparing $x + y + 1 = 0$ with $x + \alpha y + \beta = 0$,we find $\alpha = 1$ and $\beta = 1$.Therefore,$\alpha + \beta = 1 + 1 = 2$.

If the line $x + \alpha y + \beta = 0$ touches the curve $4x^3 + 4y^3 = xy(xy + 16)$ at points $(x_1, y_1)$ and $(x_2, y_2)$,where $x_1 \neq x_2$,then the value of $\alpha + \beta$ is (given $\alpha, \beta \in R$).

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