Which of the points (x,y) does not lie on the curve x^2-y^2=4 ? Solution The giv

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The given curve is --- x² + y² = 8 x y => x² + y² - 8 x y = 0 Differentiating with respect to x -- => 2 x + 2 y .y' - 8 ( 1 . y + x . y' ) = 0 where y' is the first derivative. => y' ( 2 y - 8 x ) + 2 x - 8 y = 0 => y' ( 2 y - 8 x ) = 8 y - 2 x => y ' = ( 8 y - 2 x ) / ( 2 y - 8 x ) = ( 4 y - x ) / ( y - 4 x ) Y' gives slope of the tangent ( m ) . Now slope ( m ) at the point ( 4,4 ) = 12 / ( - 12 ) = - 1 Hence its equation is -- => y = ( - 1 ) x + C .......... (1) But this tangent passes through the point ( 4,4 ). Hence 4 = ( - 1 ) ( 4 ) + C => c = 8 . Substitute C = 8 in eqn (1) to get your Answer --- y = ( - 1 ) x + 8 ------------------------- Answer ------------------------ The required tangent is x + y = 8 Note that the point ( 4,4 ) does not lie on the curve. Your question is faulty. You should have used the term " through " instead of at.

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