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(A) The curve is $y = \sqrt{x}$,which implies $y^2 = x$ for $y \ge 0$.The slope of the tangent to the curve at any point $(x, y)$ is $\frac{dy}{dx} =

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$4x + y - 18 = 0$

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(A) The curve is $y = \sqrt{x}$,which implies $y^2 = x$ for $y \ge 0$.The slope of the tangent to the curve at any point $(x, y)$ is $\frac{dy}{dx} = \frac{1}{2\sqrt{x}} = \frac{1}{2y}$.The slope of the normal at a point $(x_0, y_0)$ on the curve is $m = -\frac{1}{(dy/dx)} = -2y_0$.Since the normal passes through $(3, 6)$,the slope is $m = \frac{y_0 - 6}{x_0 - 3}$.Equating the slopes: $-2y_0 = \frac{y_0 - 6}{y_0^2 - 3}$ (since $x_0 = y_0^2$).$-2y_0(y_0^2 - 3) = y_0 - 6$ $\Rightarrow -2y_0^3 + 6y_0 = y_0 - 6$ $\Rightarrow 2y_0^3 - 5y_0 - 6 = 0$.Testing for roots,$y_0 = 2$ satisfies the equation: $2(8) - 5(2) - 6 = 16 - 10 - 6 = 0$.For $y_0 = 2$,the point on the curve is $(4, 2)$ and the slope of the normal is $m = -2(2) = -4$.The equation of the line is $y - 6 = -4(x - 3)$ $\Rightarrow y - 6 = -4x + 12$ $\Rightarrow 4x + y - 18 = 0$.

The equation of a straight line passing through the point $(3, 6)$ and cutting the curve $y = \sqrt{x}$ orthogonally is

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