Find the equation of tangent & normal to curve $2 x^{3} + 2 y^{3} - 9 x y = 0$ at the point $(2, 1)$

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Solution

$2 x^{3} + 2 y^{3} - 9 x y = 0$
$6 x^{2} + 6 y^{2} \frac{d y}{d x} - 9 y - 9 x \frac{d y}{d x} = 0$
$\Rightarrow (6 x^{2} - 9 y) + (6 y^{2} - 9 x) \frac{d y}{d x} = 0$
$\Rightarrow \frac{d y}{d x} = \frac{9 y - 6 x^{2}}{6 y^{2} - 9 x}$
At $(2, 1) \frac{d y}{d x} = \frac{9 - 24}{6 - 18} = \frac{15}{12} = \frac{5}{4}$
Equation of tangent $= (y - 1) = \frac{5}{4} (x - 2)$
$\Rightarrow 5 x - 4 y - 2 = 0$
Equation of normal $\Rightarrow (y - 1) = \frac{- 4}{5} (x - 2)$
$\Rightarrow 5 y + 4 x - 13 = 0$

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