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

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(C) Given the curve equation is $y^{2}=2(x-3)$  ...$(i)$On differentiating with respect to $x$,we get $2y \frac{dy}{dx} = 2$,which implies $\frac{dy}{

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$(5,-2)$

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(C) Given the curve equation is $y^{2}=2(x-3)$  ...$(i)$On 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 at $(x, y)$ 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 $m = 2$.Since the normal is parallel to the line,their slopes must be equal:$-y = 2 \Rightarrow y = -2$.Substituting $y = -2$ into the curve equation $(i)$:$(-2)^{2} = 2(x - 3)$$4 = 2(x - 3)$$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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