The point on the curve y^2=2(x-3) at which th | vedclass

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(D) Given curve is $y^2=2(x-3)$.Differentiating with respect to $x$,we get $2y \frac{dy}{dx} = 2$,which implies $\frac{dy}{dx} = \frac{1}{y}$.The slop

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$(5,-2)$

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(D) Given curve is $y^2=2(x-3)$.Differentiating with respect to $x$,we get $2y \frac{dy}{dx} = 2$,which implies $\frac{dy}{dx} = \frac{1}{y}$.The slope of the tangent at any point $(x, y)$ is $m_t = \frac{1}{y}$.The slope of the normal is $m_n = -\frac{1}{m_t} = -y$.The given line is $y - 2x + 1 = 0$,which can be written as $y = 2x - 1$. The slope of this line is $2$.Since the normal is parallel to the line,their slopes must be equal,so $-y = 2$,which gives $y = -2$.Substituting $y = -2$ into the curve equation: $(-2)^2 = 2(x - 3) \Rightarrow 4 = 2(x - 3) \Rightarrow 2 = x - 3 \Rightarrow x = 5$.Thus,the required point is $(5, -2)$.

The point on the curve $y^2=2(x-3)$ at which the normal is parallel to the line $y-2x+1=0$ is

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