Question

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

Answer 1

Answer: 170

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

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