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The point of the curve $y^{2}=2(x-3)$ at which the normal is parallel to the line $y-2 x+1=0$ is
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The correct answer is:
$(5,-2)$
Given, $y^{2}=2(x-3) \quad$...(i)
On differentiating w.r.t. $x$, we get $2 y \frac{d y}{d x}=2$
$\Rightarrow$
$\frac{d y}{d x}=\frac{1}{y}$
Slope of the normal $=\frac{-1}{(d y / d x)}=-y$ Slope of the given line $=2$ $\therefore \quad y=-2$
From Eq. (i), $x=5$
$\therefore$ Required point is $(5,-2)$.
On differentiating w.r.t. $x$, we get $2 y \frac{d y}{d x}=2$
$\Rightarrow$
$\frac{d y}{d x}=\frac{1}{y}$
Slope of the normal $=\frac{-1}{(d y / d x)}=-y$ Slope of the given line $=2$ $\therefore \quad y=-2$
From Eq. (i), $x=5$
$\therefore$ Required point is $(5,-2)$.
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