Question

Let the area enclosed by the $x$-axis, and the tangent and normal drawn to the curve $4 x^3-3 x y^2+6 x^2-5 x y-8 y^2+9 x+14=0$ at the point $(-2,3)$ be $A$. Then $8 A$ is equal to ______

Answer 1

$ 4 x^3-3 x y^2+6 x^2-5 x y-8 y^2+9 x+14=0 $ at $ P(-2,3)$
$ 12 x^2-3\left(y^2+2 y x y^{\prime}\right)+12 x-5\left(x y^{\prime}+y\right)-16 y y^{\prime}+ 9=0$
$ 48-3\left(9-12 y^{\prime}\right)-24-5\left(-2 y^{\prime}+3\right)-48 y^{\prime}+9 =0$
$ y^{\prime}=-9 / 2 $
Tangent $y-3=-\frac{9}{2}(x+2) \Rightarrow 9 x+2 y=-12$
Normal $: y-3=\frac{2}{9}(x+2) \Rightarrow 9 y-2 x=31$

Area $=\frac{1}{2}\left(\frac{31}{2}-4\right) \times 3=\frac{85}{4}$
$ 8 A =170$