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

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

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$(5, -2)$

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(C) Given the curve equation is ${y^2} = 2(x - 3)$ .....$(i)$Differentiating with respect to $x$,we get $2y \cdot \frac{dy}{dx} = 2$,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:$-y = 2 \implies y = -2$.Substituting $y = -2$ into equation $(i)$:$(-2)^2 = 2(x - 3)$$4 = 2(x - 3)$$2 = x - 3$$x = 5$.Therefore,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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