Normal to the curve $y = x^{3} - 2 x^{2} + 4$ at the point where $x = 2$

3 x + 4 y = 18

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x + 4 y = 18

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4 x + 3 y = 18

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4 x + y = 18

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Solution

The correct option is C $x + 4 y = 18$
Substituting $x = 2$ in the equation, we get
$y = 8 - 8 + 4 = 4$
Thus the point is $(2, 4)$ .
Now
$d y = 3 x^{2} - 4 x . d x$
Or
$\frac{- d x}{d y} = \frac{- 1}{3 x^{2} - 4 x}$
Now
${\frac{- d x}{d y}}_{x = 2} = \frac{- 1}{12 - 8}$
$= \frac{1}{4}$ .
Hence equation of normal will be
$y - 4 = (x - 2) . \frac{- 1}{4}$
Or
$4 y - 16 = 2 - x$
Or
$x + 4 y = 18$

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