Given curve

$4x^3-3xy^2+6x^2-5xy-8y^2+9x+14=0$

We need the tangent and normal at the point $(-2,3)$, and then the area enclosed by these two lines and the $x$-axis.

Differentiate implicitly

Differentiate both sides w.r.t. $x$:

$\frac{d}{dx}(4x^3)-\frac{d}{dx}(3xy^2)+\frac{d}{dx}(6x^2)-\frac{d}{dx}(5xy)-\frac{d}{dx}(8y^2)+\frac{d}{dx}(9x)=0$

$12x^2-3\left(y^2+2xy\frac{dy}{dx}\right)+12x-5\left(y+x\frac{dy}{dx}\right)-16y\frac{dy}{dx}+9=0$

Simplifying:

$12x^2-3y^2-6xy\frac{dy}{dx}+12x-5y-5x\frac{dy}{dx}-16y\frac{dy}{dx}+9=0$

Group the terms containing $\dfrac{dy}{dx}$:

$(-6xy-5x-16y)\frac{dy}{dx}+(12x^2-3y^2+12x-5y+9)=0$

Hence,

$\frac{dy}{dx}=\frac{-(12x^2-3y^2+12x-5y+9)}{-6xy-5x-16y}$

or

$\frac{dy}{dx}=\frac{12x^2-3y^2+12x-5y+9}{6xy+5x+16y}$

Slope at $(-2,3)$

Substitute $x=-2$, $y=3$:

Numerator: $12(4)-3(9)+12(-2)-5(3)+9=48-27-24-15+9=-9$

Denominator: $6(-2)(3)+5(-2)+16(3)=-36-10+48=2$

Therefore,

$m_{\text{tangent}}=\left.\frac{dy}{dx}\right|_{(-2,3)}=\frac{-9}{2}$

Equation of tangent

Using point-slope form through $(-2,3)$:

$y-3=-\frac{9}{2}(x+2)$

$2y-6=-9x-18$

$9x+2y+12=0$

Equation of normal

Slope of normal is the negative reciprocal of $-\dfrac{9}{2}$:

$m_{\text{normal}}=\frac{2}{9}$

So,

$y-3=\frac{2}{9}(x+2)$

$9y-27=2x+4$

$2x-9y+31=0$

Find x-intercepts

Since the enclosed area is with the $x$-axis, we need where these lines meet the $x$-axis.

• For the tangent, put $y=0$ in $9x+2y+12=0$: $9x+12=0 \Rightarrow x=-\frac{4}{3}$ So tangent meets the $x$-axis at $\left(-\frac{4}{3},0\right)$

• For the normal, put $y=0$ in $2x-9y+31=0$: $2x+31=0 \Rightarrow x=-\frac{31}{2}$ So normal meets the $x$-axis at $\left(-\frac{31}{2},0\right)$

Form the triangle

The tangent and normal intersect at $(-2,3)$, and together with the segment of the $x$-axis between their intercepts, they form a triangle.

Base on the $x$-axis:

$\left| -\frac{4}{3}-\left(-\frac{31}{2}\right)\right|=\left| -\frac{4}{3}+\frac{31}{2}\right|$

$=\frac{93-8}{6}=\frac{85}{6}$

Height = perpendicular distance from $(-2,3)$ to the $x$-axis = $3$.

Therefore area,

$A=\frac{1}{2}\times \frac{85}{6}\times 3=\frac{85}{4}$

Compute $8A$

$8A=8\times \frac{85}{4}=170$

Comparison with stored answer

Derived answer is $170$, which matches the stored correct answer.