Choose the correct option from the given alternatives : If the tangent at (1, 1) on y2 = x(2 – x)2 meets the curve again at P, then P is

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KollegeApply · Accepted Answer

($\frac{9}{4}$, $\frac{3}{8}$) [Hint : y2 = x(2 – x)2 = x(4 – 4x + x2) = x3 – 4x2 – 4x $\therefore$ 2ydy/dx = 3x^2 - 8x + 4 $\therefore$ dy/dx = (3x^2 - 8x + 4)/(2y) $\therefore$ `(dy/dx)_("at" (1, 1)) = (3(1)^2 - 8(1) + 4)/(2(1)) = (1)/(2)` = slope of the tangent at (1, 1) $\therefore$ equation of the tangent at (1, 1) is y – 1 = -(1)/(2)(x - 1) $\therefore$ 2y – 2 = – x + 1 $\therefore$ x + 2y = 3 Only the coordinates ($\frac{9}{4}$, $\frac{3}{8}$) satisfy both the equations y2 = x(2 – x)2 and x + 2y = 3 $\therefore$ P is ($\frac{9}{4}$, $\frac{3}{8}$)].

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