The equation of the normal to the curve $y = x (2 - x)$ at the point $(2, 0)$ is

x - 2 y = 2

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x - 2 y + 2 = 0

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2 x + y = 4

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2 x + y - 4 = 0

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Solution

The correct option is D $x - 2 y = 2$
The given curve is
$y = 2 x - x^{2}$
i.e, $x^{2} - 2 x + y = 0$

Now the equation of the tangent at the point ( $x_{1}, y_{1}$ )= $(2, 0)$
or, $x x_{1} - (x + x_{1}) + \frac{1}{2} (y + y_{1}) = 0$
i.e $2 x + (- 1) (x + 2) + \frac{1}{2} (y + 0) = 0$
i.e, $2 x - 4 + y = 0$
i.e, $y = - 2 x + 4$

So, the slope of the tangent is $(- 2)$

Therefore the slope of the normal= $\frac{1}{2}$

So,the equation of the normal at the point $(2, 0)$ is

$y - 0 = \frac{1}{2} (x - 2)$
i.e, $2 y = x - 2$
i.e, $x - 2 y = 2$

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x^{2} + y^{2} - 2 x + 4 y = 0

List-I	List-II
a) Equation of tangent to the curve $y = b e^{- x / a}$ at $x = 0$	1) $x - 2 y = 2$
b) Equation of tangent to the curve $y = x^{2} + 1$ at $(1, 2)$	2) $y = 2 x$
c) Equation of normal to the curve $y = 2 x - x^{2}$ at $(2, 0)$	3) $x - y = π$
d) Equation of normal to the curve $y = sin x$ at $x = π$	4) $\frac{x}{a} + \frac{y}{b} = 1$

(x^{2} + x y + 4 x + 2 y + 4) \frac{d y}{d x} - y^{2} = 0

y = y (x)

(x, y)

\frac{2 y}{x^{2}}

x^{2} + y^{2} - 2 x - 2 y = 0

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