Let the area enclosed by the x -axis, and the tangent and normal...

The given curve is $4{x}^{3} - 3x{y}^{2} + 6{x}^{2} - 5xy - 8{y}^{2} + 9x + 14 = 0$

On differentiating both sides, we get

$12{x}^{2} - 3{y}^{2} - 6xy{y}^{'} + 12x - 5y - 5x{y}^{'} - 16y{y}^{'} + 9 = 0$

Now, at the point \left(- 2, 3\right)$</p><p>$48 - 27 + 36{y}^{'} - 24 - 15 + 10{y}^{'} - 48{y}^{'} + 9 = 0$</p><p>i.e.${\left({y}^{'}\right)}_{\left(- 2, 3\right)} = \dfrac{- 9}{2}$</p><p>So slope of tangent${m}_{T} = \dfrac{- 9}{2}$and </p><p>slope of normal${m}_{N} = \dfrac{2}{9}$</p><p>i.e. equation of tangent$T ≡ y - 3 = \dfrac{- 9}{2}\left(x + 2\right) ⇒ y = - \dfrac{9}{2}x - 6$</p><p>and equation of normal$N ≡ y - 3 = \dfrac{2}{9}\left(x + 2\right) ⇒ y = \dfrac{2}{9}x + \dfrac{31}{9}$</p><p>Intersection point of tangent and normal is \left(- 2, 3\right)$

The intersection point of tangent and normal with $x$-axis respectively will be \left(- \dfrac{4}{3}, 0\right) \text{ and } \left(- \dfrac{31}{2}, 0\right)$</p><p>Now, area of triangle formed$= \dfrac{1}{2} \times$Base$\times$Height</p><p>$A = \dfrac{1}{2} \times \left(\dfrac{- 4}{3} + \dfrac{31}{2}\right)\left(3\right) = \dfrac{1}{2}\left(\dfrac{85}{6}\right) \cdot 3 = \dfrac{85}{4}$</p><p>$⇒ 8A = 170