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(B) The given line is $yy' = 2ax + 2ax'$,which can be rewritten as $2ax - yy' + 2ax' = 0$.Comparing this with the standard form $Ax + By + C = 0$,the

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$xy' + 2ay - 2ay' - x'y' = 0$

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(B) The given line is $yy' = 2ax + 2ax'$,which can be rewritten as $2ax - yy' + 2ax' = 0$.Comparing this with the standard form $Ax + By + C = 0$,the slope $m_1 = \frac{2a}{y'}$.Since the required line is perpendicular to this line,its slope $m_2$ must satisfy $m_1 	imes m_2 = -1$.Thus,$m_2 = -\frac{y'}{2a}$.The equation of the line passing through $(x', y')$ with slope $m_2$ is $y - y' = -\frac{y'}{2a}(x - x')$.Multiplying by $2a$,we get $2ay - 2ay' = -y'(x - x')$.$2ay - 2ay' = -xy' + x'y'$.Rearranging the terms,we get $xy' + 2ay - 2ay' - x'y' = 0$.

The equation of the line passing through the point $(x', y')$ and perpendicular to the line $yy' = 2a(x + x')$ is

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