Find the equation of the normal to the curve $x^{2} + 2 y^{2} - 4 x - 6 y + 8 = 0$ at the point whose absciassa is $2$ .

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Solution

Abscissa $= x$ coordinate $= 2$

$x^{2} + 2 y^{2} - 4 x - 6 y + 8 = 0$ ---(1)

Differentiating both sides with respect to x,

$\Rightarrow 2 x + 4 y \frac{d y}{d x} - 4 - 6 \frac{d y}{d x} = 0$

$\Rightarrow \frac{d y}{d x} = \frac{4 - 2 x}{4 y - 6} = \frac{2 - x}{2 y - 3} =$ slope of tangent

substituting $x = 2$ in eq(1), we get

$4 + 2 y^{2} - 8 - 6 y + 8 = 0$

$\Rightarrow 2 y^{2} - 6 y + 4 = 0$

$\Rightarrow y^{2} - 3 y + 2 = 0$

$\Rightarrow (y - 1) (y - 2) = 0 \Rightarrow$ $y = 1 o r y = 2$

case 1 :- $y = 1 (\frac{d y}{d x})_{(2, 1)} = \frac{0}{- 1} = 0$

eq of normal $\Rightarrow (y - 1) = \frac{- 1}{0} (x - 2) \Rightarrow x - 2 = 0$

$\Rightarrow x = 2$

case 2 :- y=2 $(\frac{d y}{d x})_{2, 2} = \frac{0}{1} = 0$

eq of normal $\Rightarrow (y - 2) = \frac{- 1}{0} (x - 2) \Rightarrow (x - 2) = 0$

$\Rightarrow x = 2$

Therefore, the equation of normal is $x = 2.$

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